25.6 - Solved Examples on Variables Separable

In the previous section, we saw how to solve first order first degree differential equations with variables separable. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 25.36
Find the particular solution of the differential equation $\cos\left(\frac{dy}{dx} \right)~=~a~~(a \in R)$ given that y = 2 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\left(\frac{dy}{dx} \right)}    & {~=~}    &{a~~(a \in R)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\cos^{-1}a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\cos^{-1}a\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[\cos^{-1}a \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{x~\cos^{-1}a~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{y}    & {~=~}    &{x~\cos^{-1}a~+~\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{y~=~x~\cos^{-1}a~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 2 when x = 0

• So substituting x = 0 and y = 2 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2}    & {~=~}    &{(0)~\cos^{-1}a~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2}    & {~=~}    &{0~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{x~\cos^{-1}a~+~2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x~\cos^{-1}a}    & {~=~}    &{y~-~2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\cos^{-1}a}    & {~=~}    &{\frac{y-2}{x}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\cos\left(\frac{y-2}{x} \right)}    & {~=~}    &{a}    \\
\end{array}}$ 

Solved example 25.37
Find the equation of the curve passing through the point (1,1) whose differential equation is $x dy~=~(2x^2 + 1) dx~(x \ne 0)$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{dy}    & {~=~}    &{\left[\frac{2x^2 + 1}{x} \right]dx}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\left[2x + \frac{1}{x} \right] dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[2x + \frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{x^2~+~\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{x^2~+~\log \left|x \right|~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{x^2~+~\log \left|x \right|~+~\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{y~=~x^2~+~\log \left|x \right|~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 1 when x = 1

• So substituting x = 1 and y = 1 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{x^2~+~\log \left|x \right|~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{1}    & {~=~}    &{(1)^2~+~\log \left|1 \right|~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{1}    & {~=~}    &{1~+~0~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{0}    \\
\end{array}}$

5. So the particular solution is:

$\small{y~=~x^2~+~\log \left|x \right|}$

6. The red curve in fig.25.19 below is the graph of this particular solution. Note that, the point (1,1) falls on the red curve only.

Fig.25.19

    ♦ For the yellow curve, C = 2  
    ♦ For the red curve, C = 0
    ♦ For the green curve, C = -0.5

Solved example 25.38
Find the equation of the curve passing through the point (0,0) and whose differential equation is $y'~=~e^x\,\sin x$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y'}    & {~=~}    &{e^x\,\sin x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{e^x\,\sin x}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\left[e^x\,\sin x \right]dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[e^x\,\sin x \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{y}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{x^2~+~\log \left|x \right|~+~\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{y~=~\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes zero when x = zero

• So substituting x = 0 and y = 0 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}e^0 (\sin 0 ~-~\cos 0)~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}(1) (0 ~-~1)~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{0}    & {~=~}    &{\frac{1}{2}(1) (-1)~+~\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{\frac{1}{2}}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{1}{2}e^x (\sin x ~-~\cos x)~+~\frac{1}{2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{2y}    & {~=~}    &{e^x (\sin x ~-~\cos x)~+~1}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{2y~-~1}    & {~=~}    &{e^x (\sin x ~-~\cos x)}    \\
\end{array}}$

Solved example 25.39
For the differential equation $xy \frac{dy}{dx}~=~(x+2)(y+2)$, find the solution curve passing through the point (1,−1).
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{xy\frac{dy}{dx}}    & {~=~}    &{(x+2)(y+2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{y}{y+2}\,dy}    & {~=~}    &{\frac{x+2}{x}\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{y}{y+2}\right]dy}}    & {~=~}    &{\int{\left[\frac{x+2}{x} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~-~2\log \left|y+2 \right|~+~\rm{C}_1}    & {~=~}    &{x~+~2\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{x~-~y~+~2\log \left|y+2 \right|~+~\log \left|x \right|}    & {~=~}    &{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{x~-~y~+~\log \left(y+2 \right)^2~+~\log \left(x \right)^2}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes −1 when x = 1

• So substituting x = 1 and y = −1 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{1~-~(-1)~+~\log \left[\left(-1+2 \right)^2 \,\left(1 \right)^2\right]}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{2~+~\log \left[\left(1 \right)^2 \,\left(1 \right)^2\right]}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{2~+~\log \left[1\right]}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{2~+~0}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{2}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x~-~y~+~\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    & {~=~}    &{2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{y-x+2}    & {~=~}    &{\log \left[\left(y+2 \right)^2 \,\left(x \right)^2\right]}    \\
\end{array}}$

Solved example 25.40
Find the equation of a curve passing through the point (−2,3), given that the slope of the tangent to the curve at any point (x,y) is $\frac{2x}{y^2}$.
Solution:
1. We know that, slope of the tangent is same as the derivative. So we can write the differential equation:
$\frac{dy}{dx}~=~\frac{2x}{y^2}$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{2x}{y^2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y^2\,dy}    & {~=~}    &{2x\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[y^2\right]dy}}    & {~=~}    &{\int{\left[2x \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{y^3}{3}~+~\rm{C}_1}    & {~=~}    &{x^2~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{x^2~-~\frac{y^3}{3}}    & {~=~}    &{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{x^2~-~\frac{y^3}{3}}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{x^2~-~\frac{y^3}{3}~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 3 when x = −2

• So substituting x = −2 and y = 3 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~\frac{y^3}{3}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{(-2)^2~-~\frac{(3)^3}{3}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{4~-~9}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{-5}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~\frac{y^3}{3}}    & {~=~}    &{-5}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{\frac{y^3}{3}}    & {~=~}    &{x^2~+~5}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{y^3}    & {~=~}    &{3x^2~+~15}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{y}    & {~=~}    &{\left(3x^2~+~15 \right)^{\frac{1}{3}}}    \\
\end{array}}$

Solved example 25.41
Find the equation of a curve passing through the point (0,−2), given that at any point (x,y) on the curve, the product of the slope of it's tangent and y-coordinate of the point is equal to the x-coordinate of the point.
Solution:
1. We know that, slope of the tangent is same as the derivative. So we can write the differential equation:
$y\,\frac{dy}{dx}~=~x$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y\,\frac{dy}{dx}}    & {~=~}    &{x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y\,dy}    & {~=~}    &{x\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[y\right]dy}}    & {~=~}    &{\int{\left[x \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{y^2}{2}~+~\rm{C}_1}    & {~=~}    &{\frac{x^2}{2}~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{\frac{x^2}{2}~-~\frac{y^2}{2}}    & {~=~}    &{\rm{C}_1~-~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{x^2~-~y^2}    & {~=~}    &{2\rm{C}_1~-~2\rm{C}_2}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{x^2~-~y^2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{x^2~-~y^2~=~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes −2 when x = 0

• So substituting x = 0 and y = −2 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~y^2}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{(0)^2~-~(-2)^2}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{0~-~4}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{-4}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~-~y^2}    & {~=~}    &{-4}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{y^2~-~x^2}    & {~=~}    &{4}    \\
\end{array}}$

Solved example 25.42
In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself?
Solution:
1. Let P be the principal amount (in Rs), at any time t (in years).
• Then from the given data, we can write:
$\frac{dP}{dt}~=~P\left(\frac{5}{100} \right)~=~0.05P$
• The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dP}{dt}}    & {~=~}    &{0.05P}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dP}{P}}    & {~=~}    &{0.05\,dt}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{P} \right]dP}}    & {~=~}    &{\int{\left[0.05 \right]dt}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\log P~+~\rm{C}_1}    & {~=~}    &{0.05t~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{\log P}    & {~=~}    &{0.05t~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{\log P}    & {~=~}    &{0.05t~+~\rm{C}_3}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{P}    & {~=~}    &{e^{0.05t + \rm{C}_3}~=~e^{0.05t}.\,e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}&{P}    & {~=~}    &{e^{0.05t}.\,\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{P~=~\rm{C}\,e^{0.05t}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• At the instant when 1000 Rs is deposited, time is zero.
• We want that particular solution in which P becomes 1000 when t = 0

• So substituting t = 0 and P = 1000 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}\,e^{0.05(0)}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}\,e^0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{1000}    & {~=~}    &{\rm{C}(1)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\rm{C}}    & {~=~}    &{1000}    \\
\end{array}}$

5. So the particular solution is:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{P}    & {~=~}    &{1000\,e^{0.05t}}    \\
\end{array}}$

6. We want the time when Rs. 1000/- become Rs. 2000/-
So based on the particular solution, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{P}    & {~=~}    &{\rm{C}\,e^{0.05t}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}        &{2000}    & {~=~}    &{1000\,e^{0.05t}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}        &{2}    & {~=~}    &{e^{0.05t}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}        &{\log 2}    & {~=~}    &{0.05t}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{\log 2}    & {~=~}    &{(1/20)t}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}        &{t}    & {~=~}    &{20 \log_e 2}    \\
\end{array}}$

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The link below gives a few more solved examples:

Exercise 9.4

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In the next section, we will see homogeneous differential equations.

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25.5 - Solving Differential Equations With Variables Separable

In the previous section, we completed a discussion on formation of a differential equation whose general solution is given. In this section, we will see methods for solving first order, first degree differential equations.

• Note that, the differential equations that we consider for the present discussion will be of:
    ♦ first order
    ♦ first degree
(See section 25.2)

First we will solve those differential equations, whose variables are separable. It can be explained using an example, in 4 steps:

1. Consider the differential equation: $\small{\frac{dy}{dx}~=~\frac{x+1}{2-y},~(y \ne 2)}$
• Here, the derivative is the product of two functions: h(y) and g(x).
• We can write:
$\small{\frac{dy}{dx}~=~h(y).g(x)~=~\frac{x+1}{2-y}}$  

Where,
    ♦ $\small{g(x)~=~x+1}$
    ♦ $\small{h(y)~=~\frac{1}{2-y}}$  

2. Now the variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+1}{2-y}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{(2-y)\,dy}    & {~=~}    &{(x+1)\,dx}    \\
\end{array}}$

3. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[2-y \right]dy}}    & {~=~}    &{\int{\left[x+1 \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y~-~\frac{y^2}{2}~+~\rm{C}_1}    & {~=~}    &{\frac{x^2}{2}~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{4y~-~y^2~+~\rm{2C}_1}    & {~=~}    &{x^2~+~2\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{0}    & {~=~}    &{x^2~+~y^2~-~4y~+~2\rm{C}_2 - 2\rm{C}_1}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~4y~+~\rm{C}}    & {~=~}    &{0}    \\
\end{array}}$

4. So the general solution is:
$\small{x^2~+~y^2~-~4y~+~\rm{C}~=~0}$

---

Now we will see some solved examples

Solved example 25.25
Find the general solution of the differential equation $\frac{dy}{dx}~=~\frac{1 + y^2}{1 + x^2}$.
Solution:
1. For any real numbers x, y, the quantities $\small{1 + y^2}$ and $\small{1 + x^2}$ will not be zero. So the variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1+y^2}{1+x^2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{1+y^2}}    & {~=~}    &{\frac{dx}{1+x^2}}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{1+y^2}\right]dy}}    & {~=~}    &{\int{\left[\frac{1}{1+x^2} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\tan^{-1} y~+~\rm{C}_1}    & {~=~}    &{\tan^{-1}x~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\tan^{-1} y}    & {~=~}    &{\tan^{-1}x~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\tan^{-1} y}    & {~=~}    &{\tan^{-1}x~+~\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{\tan^{-1}y~=~\tan^{-1}x~+~\rm{C}}$

Solved example 25.26
Find the general solution of the differential equation $\frac{dy}{dx}~=~\frac{1 - \cos x}{1 + \cos x}$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{1 - \cos x}{1 + \cos x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\left[\frac{1 - \cos x}{1 + \cos x} \right]dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[\frac{1 - \cos x}{1 + \cos x} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{\int{\left[\tan^{2}\left(\frac{x}{2} \right) \right]dx}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{\tan\left(\frac{x}{2} \right)~-~x~+~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{\tan\left(\frac{x}{2} \right)~-~x~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{\tan\left(\frac{x}{2} \right)~-~x~+~\rm{C}}    \\
\end{array}}$

The reader may write all the steps for the integration in [(2) magenta color]

3. So the general solution is:
$\small{y~=~\tan\left(\frac{x}{2} \right)~-~x~+~\rm{C}}$

Solved example 25.27
Find the general solution of the differential equation $\frac{dy}{dx}~=~\sqrt{4 - y^2}~(-2 < y < 2)$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\sqrt{4 - y^2}~~(-2 < y < 2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{\sqrt{4 - y^2}}}    & {~=~}    &{dx}    \\
\end{array}}$

Since (−2 < y < 2), the denominator in the L.H.S will never become zero.

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{\sqrt{4 - y^2}}\right]dy}}    & {~=~}    &{\int{\left[1 \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\tan^{-1}\left(\frac{y}{\sqrt{4 - y^2}} \right)~+~\rm{C}_1}    & {~=~}    &{x~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\tan^{-1}\left(\frac{y}{\sqrt{4 - y^2}} \right)}    & {~=~}    &{x~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\tan(x+\rm{C})}    & {~=~}    &{\frac{y}{\sqrt{4 - y^2}}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\tan^2(x+\rm{C})}    & {~=~}    &{\frac{y^2}{4 - y^2}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\cot^2(x+\rm{C})}    & {~=~}    &{\frac{4 - y^2}{y^2}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\cot^2(x+\rm{C})}    & {~=~}    &{\frac{4}{y^2}~-~1}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{1~+~\cot^2(x+\rm{C})}    & {~=~}    &{\frac{4}{y^2}}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}    &{\csc^2(x+\rm{C})}    & {~=~}    &{\frac{4}{y^2}}    \\
{~\color{magenta}    10    }    &{{\Rightarrow}}    &{\sin^2(x+\rm{C})}    & {~=~}    &{\frac{y^2}{4}}    \\
{~\color{magenta}    11    }    &{{\Rightarrow}}    &{y^2}    & {~=~}    &{4 \sin^2(x+\rm{C})}    \\
{~\color{magenta}    12    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{2 \sin(x+\rm{C})}    \\
\end{array}}$

The reader may write all the steps for the integration in the L.H.S of [(1) magenta color]

◼ Remarks:
• In [(12) magenta color], we need not consider the −ve root. The reason can be written in two steps:
(i) What ever be the value of C, the value of $\sin(x+\rm{C})$ will lie between −1 and 1
(ii) So the value of $2 \sin(x+\rm{C})$ will lie between −2 and 2
(iii) Likewise, the value of $−2 \sin(x+\rm{C})$ will also lie between −2 and 2
(iv) We are already given that, −2 < y < 2. So there is no need to consider the −ve root.

3. So the general solution is:
$\small{y~=~2 \sin(x+\rm{C})}$

Solved example 25.28
Find the general solution of the differential equation $\frac{dy}{dx}~+~y~=~1~~(y \ne 1)$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}~+~y}    & {~=~}    &{1~~(y \ne 1)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{1-y}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{dy}{1-y}}    & {~=~}    &{dx}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{y-1}}    & {~=~}    &{(-1)dx}    \\
\end{array}}$

Since $y \ne 1$, the denominator in the L.H.S will never become zero.

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{y-1}\right]dy}}    & {~=~}    &{\int{\left[-1 \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\log(y-1)~+~\rm{C}_1}    & {~=~}    &{-x~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\log(y-1)}    & {~=~}    &{-x~+~\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\log(y-1)}    & {~=~}    &{-x~+~\rm{C}_3}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y-1}    & {~=~}    &{e^{\left(\rm{C}_3 - x \right)}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{1~+~e^{\left(\rm{C}_3 - x \right)}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{1~+~\left[e^{\rm{C}_3 }.e^{- x } \right]}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{y}    & {~=~}    &{1~+~\rm{C}\,e^{- x } }    \\
\end{array}}$

3. So the general solution is:
$\small{y~=~1~+~\rm{C}\,e^{- x } }$

Solved example 25.29
Find the general solution of the differential equation $\sec^2 x\,\tan y\,dx~+~\sec^2 y\,\tan x\,dy~=~0$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sec^2 x\,\tan y\,dx~+~\sec^2 y\,\tan x\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\sec^2 x}{\tan x}\,dx~+~\frac{\sec^2 y}{\tan y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{\sec^2 x\,\cos x}{\sin x}\,dx~+~\frac{\sec^2 y\, \cos y}{\sin y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{\sec x}{\sin x}\,dx~+~\frac{\sec y}{\sin y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{2}{\sin(2x)}\,dx~+~\frac{2}{\sin(2y)}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{1}{\sin(2x)}\,dx~+~\frac{1}{\sin(2y)}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\frac{1}{\sin(2y)}\,dy}    & {~=~}    &{\frac{-1}{\sin(2x)}\,dx}    \\
\end{array}}$

◼ Remarks:
• In [(1) magenta color], we divide the whole equation by tan x tan y.
2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{\sin(2y)}\right]dy}}    & {~=~}    &{\int{\left[\frac{-1}{\sin(2x)} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{1}{2}\log \left|\tan y \right|~+~\rm{C}_1}    & {~=~}    &{\frac{-1}{2}\log \left|\tan x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\log \left|\tan y \right|~+~2\rm{C}_1}    & {~=~}    &{-\log \left|\tan x \right|~+~2\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\log \left|\tan y \right|~+~\log \left|\tan x \right|}    & {~=~}    &{2\left(\rm{C}_2~-~\rm{C}_1 \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\log \left|\tan x\,\tan y \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\tan x\,\tan y}    & {~=~}    &{e^{\rm{C}_3}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}&{\tan x\,\tan y}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

The reader may write all the steps for the integration in [(1) magenta color]

3. So the general solution is:
$\small{\tan x\,\tan y~=~\rm{C}}$

Solved example 25.30
Find the general solution of the differential equation $y \log y\,dx~-~x\,dy~=~0$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y \log y\,dx~-~x\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x\,dy}    & {~=~}    &{y \log y\,dx}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{dy}{y \log y}}    & {~=~}    &{\frac{dx}{x}}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{y \log y}\right]dy}}    & {~=~}    &{\int{\left[\frac{1}{x} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\log\left[\log(y) \right]~+~\rm{C}_1}    & {~=~}    &{\log(x)~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\log\left[\log(y) \right]~-~\log(x)}    & {~=~}    &{\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\log\left[\frac{\log(y)}{x} \right]}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\log(y)}{x}}    & {~=~}    &{e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\log(y)}{x}}    & {~=~}    &{\rm{C}_4}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\log(y)}    & {~=~}    &{x\,\rm{C}_4}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}&{y}    & {~=~}    &{e^{x\,\rm{C}_4}}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}&{y}    & {~=~}    &{e^{(\rm{C}\,x)}}    \\
\end{array}}$

The reader may write all the steps for the integration in [(1) magenta color]

3. So the general solution is:
$\small{y~=~e^{(\rm{C}\,x)}}$

Solved example 25.31
Find the general solution of the differential equation $x^5\,\frac{dy}{dx}~=~(-1)\,y^5$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^5\,\frac{dy}{dx}}    & {~=~}    &{(-1)\,y^5}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{(-1)\,y^5}}    & {~=~}    &{\frac{dx}{x^5}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{dy}{y^5}}    & {~=~}    &{\frac{(-1)\,dx}{x^5}}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{y^5}\right]dy}}    & {~=~}    &{\int{\left[\frac{-1}{x^5} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{-1}{4y^4}~+~\rm{C}_1}    & {~=~}    &{\frac{1}{4x^4}~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{-1}{y^4}~+~4\rm{C}_1}    & {~=~}    &{\frac{1}{x^4}~+~4\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{1}{x^4}~+~\frac{1}{y^4}}    & {~=~}    &{4\left(\rm{C}_1~-~\rm{C}_2 \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x^{-4}~+~y^{-4}}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

3. So the general solution is:
$\small{x^{-4}~+~y^{-4}~=~\rm{C}}$

Solved example 25.32
Find the general solution of the differential equation $e^x\,\tan y\,dx~+~(1 - e^x) \sec^2 y\,dy~=~0$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{e^x\,\tan y\,dx~+~(1 - e^x) \sec^2 y\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{e^x\,\tan y}{\tan y (1 - e^x)}\,dx~+~\frac{(1 - e^x) \sec^2 y}{\tan y (1 - e^x)}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{e^x}{1 - e^x}\,dx~+~\frac{\sec^2 y}{\tan y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{e^x}{1 - e^x}\,dx~+~\frac{\sec^2 y\,\cos y}{\sin y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{e^x}{1 - e^x}\,dx~+~\frac{\sec y}{\sin y}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{e^x}{1 - e^x}\,dx~+~\frac{2}{\sin (2y)}\,dy}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\frac{2}{\sin (2y)}\,dy}    & {~=~}    &{\frac{e^x}{e^x - 1}\,dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{2}{\sin (2y)}\right]dy}}    & {~=~}    &{\int{\left[\frac{e^x}{e^x - 1} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\log \left|\tan y \right|~+~\rm{C}_1}    & {~=~}    &{\log \left|e^x - 1 \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\log \left|\tan y \right|~-~\log \left|e^x - 1 \right|}    & {~=~}    &{\rm{C}_2~-~\rm{C}_1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\log \left|\frac{\tan y}{e^x - 1} \right|}    & {~=~}    &{\rm{C}_3}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{\tan y}{e^x - 1}}    & {~=~}    &{e^{\rm{C}_3}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{\tan y}{e^x - 1}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}&{\tan y}    & {~=~}    &{\rm{C}\left(e^x - 1 \right)}    \\
\end{array}}$

The reader may write all the steps for the integration in [(1) magenta color]

3. So the general solution is:
$\small{\tan y~=~\rm{C}\left(e^x - 1 \right)}$  

Solved example 25.33
Find the particular solution of the differential equation $\frac{dy}{dx}~=~-4xy^2$ given that y = 1 when x = 0.
Solution:
1. For any real number y, the quantity $\small{y^2}$ will not be zero. So the variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{-4xy^2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{y^2}}    & {~=~}    &{-4x \, dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{y^2}\right]dy}}    & {~=~}    &{\int{\left[-4x \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{-1}{y}~+~\rm{C}_1}    & {~=~}    &{-2x^2~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{2 x^2~-~\frac{1}{y}~+~\rm{C}_1~-~\rm{C}_2}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{2 x^2~-~\frac{1}{y}~+~\rm{C}}    & {~=~}    &{0}    \\
\end{array}}$

3. So the general solution is:
$\small{2 x^2~-~\frac{1}{y}~+~\rm{C}~=~0}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 1 when x = 0

• So substituting x = 0 and y = 1 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2 (0)^2~-~\frac{1}{(1)}~+~\rm{C}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\rm{C}}    & {~=~}    &{1}    \\
\end{array}}$

5. So the particular solution is:

$\small{2 x^2~-~\frac{1}{y}~+~1~=~0}$

6. The red curve in fig.25.18 below is the graph of this particular solution. Note that, the point (0,1) falls on the red curve only.

Fig.25.18

    ♦ For the yellow curve, C = 2  
    ♦ For the red curve, C = 1
    ♦ For the green curve, C = 0.75

Solved example 25.34
Find the particular solution of the differential equation $\left(x^3 + x^2 + x + 1 \right)\frac{dy}{dx}~=~2x^2~+~x$ given that y = 1 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\left(x^3 + x^2 + x + 1 \right)\frac{dy}{dx}}    & {~=~}    &{2x^2~+~x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\frac{2x^2~+~x}{x^3 + x^2 + x + 1} \, dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[\frac{2x^2~+~x}{x^3 + x^2 + x + 1} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{\frac{3}{4} \log\left(x^2 + 1 \right)~+~\frac{1}{2} \log(x+1)~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{3}{4} \log\left(x^2 + 1 \right)~+~\frac{2}{4} \log(x+1)~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{4} \log\left(x^2 + 1 \right)^3~+~\frac{1}{4} \log(x+1)^2~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}_2}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{4} \left[\log\left(x^2 + 1 \right)^3~+~\log(x+1)^2 \right]~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}_2}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{4} \log\left[\left(x^2 + 1 \right)^3~(x+1)^2 \right]~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}_2}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}&{y}    & {~=~}    &{\frac{1}{4} \log\left[\left(x^2 + 1 \right)^3~(x+1)^2 \right]~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}}    \\
\end{array}}$

The reader may write all the steps for the integration in [(1) magenta color]

3. So the general solution is:
$\small{y~=~\frac{1}{4} \log\left[\left(x^2 + 1 \right)^3~(x+1)^2 \right]~-~\frac{1}{2} \tan^{-1} x~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 1 when x = 0

• So substituting x = 0 and y = 1 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{1}    & {~=~}    &{\frac{1}{4} \log\left[\left(0^2 + 1 \right)^3~(0+1)^2 \right]~-~\frac{1}{2} \tan^{-1} 0~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{1}    & {~=~}    &{\frac{1}{4} \log\left[1 \right]~-~\frac{1}{2} (0)~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{1}    & {~=~}    &{0~-~0~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{1}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

5. So the particular solution is:

$\small{y~=~\frac{1}{4} \log\left[\left(x^2 + 1 \right)^3~(x+1)^2 \right]~-~\frac{1}{2} \tan^{-1} x~+~1}$

Solved example 25.35
Find the particular solution of the differential equation $x(x^2 - 1)\frac{dy}{dx}~=~1$ given that y = 0 when x = 2.
Solution:
1. The variables can be separated and the differential equation can be rewritten:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x(x^2 - 1)\frac{dy}{dx}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{dy}    & {~=~}    &{\frac{1}{x(x^2 - 1)} \, dx}    \\
\end{array}}$

2. Integrating both sides, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[1\right]dy}}    & {~=~}    &{\int{\left[\frac{1}{x(x^2 - 1)} \right]dx}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{2} \log \left|1-x^2 \right|~-~\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{2} \log \left|1-x^2 \right|~-~\frac{2}{2}\log \left|x \right|~+~\rm{C}_2}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{2} \left[\log \left|1-x^2 \right|~-~\log \left|x^2 \right| \right]~+~\rm{C}_2}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}&{y~+~\rm{C}_1}    & {~=~}    &{\frac{1}{2} \log \left|\frac{1-x^2}{x^2} \right|~+~\rm{C}_2}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}&{y}    & {~=~}    &{\frac{1}{2} \log \left|\frac{1-x^2}{x^2} \right|~+~\rm{C}}    \\
\end{array}}$

The reader may write all the steps for the integration in [(1) magenta color]

3. So the general solution is:
$\small{y~=~\frac{1}{2} \log \left|\frac{1-x^2}{x^2} \right|~+~\rm{C}}$

4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes zero when x = 2

• So substituting x = 2 and y = 0 in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{0}    & {~=~}    &{\frac{1}{2} \log \left|\frac{1-2^2}{2^2} \right|~+~\rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{0}    & {~=~}    &{\frac{1}{2} \log \left|\frac{-3}{4} \right|~+~\rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{0}    & {~=~}    &{\frac{1}{2} \log \left(\frac{3}{4} \right)~+~\rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{-\frac{1}{2} \log \left(\frac{3}{4} \right)}    & {~=~}    &{\rm{C}}    \\
\end{array}}$

5. So the particular solution is:

$\small{y~=~\frac{1}{2} \log \left|\frac{1-x^2}{x^2} \right|~-~\frac{1}{2} \log \left(\frac{3}{4} \right)}$

---

In the next section, we will see a few more solved examples.

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25.4 - Solved Examples Related To Formation of Differential Equation

In the previous section, we saw formation of a differential equation whose general solution is given. We saw a solved example also. In this section, we will see a few more solved examples.

Solved example 25.17
The equation $y^2~=~a(b^2 - x^2)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{a(b^2 - x^2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y\,y'}    & {~=~}    &{0~-~2x\,a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{-ax}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y\,y''~+~y'\,y'}    & {~=~}    &{-a}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y\,y''~+~(y')^2}    & {~=~}    &{\frac{y\,y'}{x}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{xy\,y''~+~x\,(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(5) magenta color], we use the result from [(3) magenta color] to eliminate 'a'

2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x\,(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

Solved example 25.18
The equation $y~=~a \sin(x+b)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{a \sin(x+b)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{a \cos (x+b)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y''}    & {~=~}    &{-a \sin(x+b)}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y''}    & {~=~}    &{-y}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y''~+~y}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate 'a' and 'b'

2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y''~+~y~=~0}$ represents the given family of curves.

Solved example 25.19
Form the differential equation representing the family of ellipses having foci on the x-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1}$ (see section 11.4)

A few members of the family are shown in the fig.25.12 below:

Fig.25.12

For the green ellipse, a = 5 and b = 3
For the red ellipse, a = 3 and b = 2
For the yellow ellipse, a = 7 and b = 4

2. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{2x}{a^2}~+~\frac{2y\,y'}{b^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{a^2}~+~\frac{y\,y'}{b^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x b^2~+~y\,y'\,a^2}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{ b^2~+~a^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'\,a^2}{x}~+~a^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{xy\,y''~+~x(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{b^2}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

Solved example 25.20
Form the differential equation representing the family of ellipses having foci on the y-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1}$

A few members of the family are shown in the fig.25.13 below:

Fig.25.13

For the green ellipse, a = 3 and b = 5
For the red ellipse, a = 2 and b = 3
For the yellow ellipse, a = 4 and b = 7

2. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{2x}{b^2}~+~\frac{2y\,y'}{a^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{b^2}~+~\frac{y\,y'}{a^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x a^2~+~y\,y'\,b^2}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{ a^2~+~b^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'\,b^2}{x}~+~b^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{xy\,y''~+~x(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{a^2}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

4. This is the same result as in the previous solved example 25.19. The reason can be written in two steps:
(i) The algebraic equation for the two types of ellipses are basically the same. The only difference is that, the arbitrary constants 'a' and 'b' are interchanged.
(ii) In the differential equation, there is no role for the arbitrary constants. So we get the same result. 

Solved example 25.21
Form the differential equation representing the family of circles touching the x-axis at the origin.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.14 below:

Fig.25.14

• Based on the fig., we can write:
'a' will be zero and 'b' will be equal to 'r'.

• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~(y-r)^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2yr~+~r^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2yr}    & {~=~}    &{0}    \\
\end{array}}$

• In the above fig.25.14,
   ♦ r = 1 for the green circle
   ♦ r = 3 for the red circle
   ♦ r = 2 for the yellow circle

2. To eliminate 'r', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2~-~2yr}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y\,y'~-~2r\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~r\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x~+~(y~-~r)\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x~+~\left[y~-~\left(\frac{x^2~+~y^2}{2y} \right)\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{x~+~\left[\frac{2y^2~-~x^2~-~y^2}{2y}\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{x~+~\left[\frac{y^2~-~x^2}{2y}\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{\left[\frac{x^2~-~y^2}{2y}\right]\,y'}    & {~=~}    &{x}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{2xy}{x^2~-~y^2}}    \\
\end{array}}$

◼ Remarks:
In [(5) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2xy}{x^2~-~y^2}}$ represents the given family of curves.

Solved example 25.22
Form the differential equation representing the family of circles touching the y-axis at the origin.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.15 below:

Fig.25.15

• Based on the fig., we can write:
'b' will be zero and 'a' will be equal to 'r'.

• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x-r)^2~+~y^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x^2~-~2xr~+~r^2~+~y^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2xr}    & {~=~}    &{0}    \\
\end{array}}$

• In the above fig.25.15,
   ♦ r = 1 for the green circle
   ♦ r = 3 for the red circle
   ♦ r = 2 for the yellow circle

2. To eliminate 'r', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2~-~2xr}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y\,y'~-~2r}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~r}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~\left(\frac{x^2~+~y^2}{2x} \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[x~-~\left(\frac{x^2~+~y^2}{2x} \right) \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\frac{2x^2~-~x^2~-~y^2}{2x} \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\left[\frac{x^2~-~y^2}{2x} \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{\left[\frac{y^2~-~x^2}{2x} \right]}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{y^2~-~x^2}{2xy}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y^2~-~x^2}{2xy}}$ represents the given family of curves.

Solved example 25.23
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{y^2~=~4ax}$

A few members of the family are shown in the fig.25.16 below:

Fig.25.16

For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3

2. To eliminate 'a', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{4ax}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y\,y'}    & {~=~}    &{4a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{2a}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{2\left(\frac{y^2}{4x} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{y}{2x}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y}{2x}}$ represents the given family of curves.

Solved example 25.24
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of y-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{x^2~=~4ay}$

A few members of the family are shown in the fig.25.17 below:

Fig.25.17

For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3

2. To eliminate 'a', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2}    & {~=~}    &{4ay}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x}    & {~=~}    &{4a\,y'}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x}    & {~=~}    &{2a\,y'}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x}    & {~=~}    &{y'\left(\frac{x^2}{2y} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{1}    & {~=~}    &{y'\left(\frac{x}{2y} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{y'}    & {~=~}    &{\frac{2y}{x}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2y}{x}}$ represents the given family of curves.

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The link below gives a few more solved examples:

Exercise 9.3

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In the next section, we will see solution of differential equations.

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