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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~-~y^2} & {~=~} &{-4} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y^2~-~x^2} & {~=~} &{4} \\
\end{array}}$
The link below gives a few more solved examples:
Exercise 9.4
In the next section, we will see homogeneous differential equations.
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