Chapter 9: Differential Equations

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{1- cos x}}{\text{1+ cos x}}$.

Use trigonometric half –angle identities to simplify:
$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$ 

$\Rightarrow \frac{dy}{dx}=\frac{2{{\sin }^{2}}\frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}$ 

$\Rightarrow \frac{dy}{dx}={{\tan }^{2}}\frac{x}{2}$ 

$\Rightarrow \frac{dy}{dx}={{\sec }^{2}}\frac{x}{2}-1$ 

Separate the differentials and integrate:

$\int dy=\int \left( {{\sec }^{2}}\frac{x}{2}-1 \right)dx$ 

$\Rightarrow y=\int {{\sec }^{2}}\frac{x}{2}dx-\int dx$ 

$\Rightarrow y=2\tan \frac{x}{2}-x+C$ 

Thus the general solution of given differential equation is $\text{y = 2tan}\frac{\text{x}}{\text{2}}\text{ - x + C}$.

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{4 - }{{\text{y}}^{\text{2}}}}\left( \text{-2  y  2} \right)$.

Simplify the expression:

$\frac{dy}{dx}=\sqrt{4-{{y}^{2}}}$ 

$\Rightarrow \frac{dy}{\sqrt{4-{{y}^{2}}}}=dx$ 

Use standard integration:

$\int \frac{dy}{\sqrt{4-{{y}^{2}}}}=\int dx$ 

$\Rightarrow {{\sin }^{-1}}\frac{y}{2}=x+C$ 

$\Rightarrow \frac{y}{2}=\sin \left( x+C \right)$ 

$\Rightarrow y=2\sin \left( x+C \right)$ 

Thus the general solution of given differential equation is $\text{y = 2 sin}\left( \text{x + C} \right)$.

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{+ y =1}\left( \text{y}\ne \text{1} \right)$.

Simplify the expression:

$\frac{dy}{dx}+y=1$ 

$\Rightarrow \frac{dy}{dx}=1-y$ 

$\Rightarrow \frac{dy}{1-y}=dx$ 

Use standard integration:

$\int \frac{dy}{1-y}=\int dx$ 

$\Rightarrow -\log \left( 1-y \right)=x+C$ 

$\Rightarrow \log \left( 1-y \right)=-\left( x+C \right)$ 

$\Rightarrow 1-y={{e}^{-\left( x+C \right)}}$ 

$y=1-A{{e}^{-x}}\,\left( A={{e}^{-C}} \right)$ 

Thus the general solution of given differential equation is $\text{y=1- A}{{\text{e}}^{\text{-x}}}\,$.

Ans: The given differential equation is:

${{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy=0$.

Divide both side by $\text{tan x tan y}$:

$\frac{{{\sec }^{2}}x\tan ydx+{{\sec }^{2}}y\tan xdy}{\tan x\tan y}=0$ 

\[\Rightarrow \frac{{{\sec }^{2}}x}{\tan x}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0\] 

Integrate both side:

$\int \frac{{{\sec }^{2}}x}{\tan x}dx+\int \frac{{{\sec }^{2}}y}{\tan y}dy=0$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{\sec }^{2}}x}{\tan x}dx$       ……(1)

Use a substitution method for integration. Substitute $\text{tan x = u}$:

For integral on RHS:

$\Rightarrow \tan x=u$ 

$\Rightarrow {{\sec }^{2}}xdx=du$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}x}{\tan x}dx=\log \left( \tan x \right)$ 

Thus evaluating result form (1):

$\Rightarrow \log \left( \tan y \right)=-\log \left( \tan x \right)+\log \left( C \right)$ 

$\Rightarrow \log \left( \tan y \right)=\log \left( \frac{C}{\tan x} \right)$ 

$\Rightarrow \tan x\tan y=C$ 

Thus the general solution of given differential equation is $\text{tan x tan y = C}$.

Ans: The given differential equation is:

$\left( {{e}^{x}}+{{e}^{-x}} \right)dy-\left( {{e}^{x}}-{{e}^{-x}} \right)dx=0$.

Simplify the expression:

$dy=\left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ 

Integrate both side:

$\int dy=\int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx$ ……(1)

Use a substitution method for integration. Substitute \[{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}\text{=t}\]:

For integral on RHS:

$\Rightarrow {{e}^{x}}+{{e}^{-x}}=t$ 

$\Rightarrow \left( {{e}^{x}}-{{e}^{-x}} \right)dx=dt$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\int \frac{dt}{t}$ s

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\ln t+C$ 

$\Rightarrow \int \left[ \frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}} \right]dx=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus evaluating result form (1):

$y=\log \left( {{e}^{x}}+{{e}^{-x}} \right)+C$ 

Thus the general solution of given differential equation is $\text{y=log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$.

Ans: The given differential equation is:

$\frac{dy}{dx}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)$.

Simplify the expression:

$\frac{dy}{1+{{y}^{2}}}=\left( 1+{{x}^{2}} \right)dx$ 

Integrate both side:

$\int \frac{dy}{1+{{y}^{2}}}=\int \left( 1+{{x}^{2}} \right)dx$

Use standard integration:

${{\tan }^{-1}}y=\int dx+\int {{x}^{2}}dx$ 

$\Rightarrow {{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{3}+C$ 

Thus the general solution of given differential equation is $\text{ta}{{\text{n}}^{\text{-1}}}\text{y=x+}\frac{{{\text{x}}^{\text{3}}}}{\text{3}}\text{+C}$.

Ans: The given differential equation is:

$y\log ydx-xdy=0$.

Simplify the expression:

$y\log ydx=xdy$ 

$\Rightarrow \frac{dx}{x}=\frac{dy}{y\log y}$ 

Integrate both side:

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute \[\text{log y=t}\]:

$\log y=t$ 

$\Rightarrow \frac{1}{y}dy=dt$ 

$\Rightarrow \int \frac{dy}{y\log y}=\int \frac{dt}{t}$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log t$ 

$\Rightarrow \int \frac{dy}{y\log y}=\log \left( \log y \right)$ 

Evaluating expression (1):

$\log \left( \log y \right)=\log x+\log C$ 

$\Rightarrow \log \left( \log y \right)=\log Cx$ 

$\Rightarrow \log y=Cx$ 

$\Rightarrow y={{e}^{Cx}}$ 

Thus the general solution of given differential equation is $\text{y=}{{\text{e}}^{\text{Cx}}}$.

Ans: The given differential equation is:

${{x}^{5}}\frac{dy}{dx}=-{{y}^{5}}$.

Simplify the expression:

$\frac{dy}{{{y}^{5}}}=-\frac{dx}{{{x}^{5}}}$ 

Integrate both side:

$\int \frac{dy}{{{y}^{5}}}=-\int \frac{dx}{{{x}^{-5}}}$ 

\[\Rightarrow \int {{y}^{-5}}dy=-\int {{x}^{-5}}dx\] 

$\Rightarrow \frac{{{y}^{-5+1}}}{-5+1}=-\frac{{{x}^{-5+1}}}{-5+1}+C$ 

$\Rightarrow \frac{{{y}^{-4}}}{-4}=-\frac{{{x}^{-4}}}{-4}+C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=-4C$ 

$\Rightarrow {{x}^{-4}}+{{y}^{-4}}=A\,\,\,\,\left( A=-4C \right)$ 

Thus the general solution of given differential equation is ${{\text{x}}^{\text{-4}}}\text{+}{{\text{y}}^{\text{-4}}}\text{=A}$.

Ans: The given differential equation is:

$\frac{dy}{dx}={{\sin }^{-1}}x$.

Simplify the expression:

$dy={{\sin }^{-1}}xdx$ 

Integrate both side:

$\int dy=\int {{\sin }^{-1}}xdx$ 

\[\Rightarrow y=\int 1\times {{\sin }^{-1}}xdx\] 

Use product rule of integration:

$\int {{\sin }^{-1}}xdx={{\sin }^{-1}}x\int dx-\int \left( \frac{1}{\sqrt{1-{{x}^{2}}}}\int dx \right)dx$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x-\int \frac{x}{\sqrt{1-{{x}^{2}}}}dx$ 

Substitute $\text{1-}{{\text{x}}^{\text{2}}}\text{=}{{\text{t}}^{\text{2}}}$ 

$1-{{x}^{2}}={{t}^{2}}$ 

$\Rightarrow -2xdx=2tdt$ 

$\Rightarrow -xdx=tdt$ 

Evaluating the integral:

$\frac{2{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\int \frac{tdt}{\sqrt{{{t}^{2}}}}$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+t+C$ 

$\Rightarrow \int {{\sin }^{-1}}xdx=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

$\Rightarrow y=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C$ 

Thus the general solution of given differential equation is $\text{y=xsi}{{\text{n}}^{\text{-1}}}\text{x+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{+C}$,

Ans: The given differential equation is:

${{e}^{x}}\tan ydx+\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=0$.

Simplify the expression:

$\left( 1-{{e}^{x}} \right){{\sec }^{2}}ydy=-{{e}^{x}}\tan ydx$ 

$\Rightarrow \frac{{{\sec }^{2}}y}{\tan y}dy=-\frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ 

Integrate both side:

$\int \frac{{{\sec }^{2}}y}{\tan y}dy=-\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx$ ……(1)

Substitute $\text{tan y=u}$ 

$\tan y=u$ 

\[\Rightarrow {{\sec }^{2}}y=du\] 

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\int \frac{du}{u}$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log u$ 

$\Rightarrow \int \frac{{{\sec }^{2}}y}{\tan y}dy=\log \left( \tan y \right)$ 

Substitute $\text{1-}{{\text{e}}^{x}}\text{=v}$ 

$1-{{e}^{x}}=v$ 

$\Rightarrow -{{e}^{x}}dx=dv$ 

Evaluating the RHS integral of (1):

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\int \frac{dv}{v}$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log v$ 

$\Rightarrow -\int \frac{{{e}^{x}}}{\left( 1-{{e}^{x}} \right)}dx=\log \left( 1-{{e}^{x}} \right)$ 

Therefore the integral (1) will be:

$\log \left( \tan y \right)=\log \left( 1-{{e}^{x}} \right)+\log C$ 

$\Rightarrow \log \left( \tan y \right)=\log C\left( 1-{{e}^{x}} \right)$ 

$\Rightarrow \tan y=C\left( 1-{{e}^{x}} \right)$ 

Thus the general solution of given differential equation is $\text{tany=C}\left( \text{1-}{{\text{e}}^{\text{x}}} \right)$,