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:
$\small{y~=~\frac{1}{2} \log \left|\frac{1-x^2}{x^2} \right|~-~\frac{1}{2} \log \left(\frac{3}{4} \right)}$
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
In the next section, we will see a few more solved examples.
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