Class 12: Differential Equations

$\displaystyle \text{Solve the following differential equations:}$

$\displaystyle \textbf{Question 1: }~\frac{dy}{dx}+\frac{1+y^{2}}{y}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{dy}{dx}+\frac{1+y^{2}}{y}=0$
$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{1+y^{2}}{y}$
$\displaystyle \Rightarrow \frac{dx}{dy}=-\frac{y}{1+y^{2}}$
$\displaystyle \Rightarrow dx=-\left(\frac{y}{1+y^{2}}\right)dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \Rightarrow \int dx=-\int\left(\frac{y}{1+y^{2}}\right)dy$
$\displaystyle \Rightarrow x=-\int\left(\frac{y}{1+y^{2}}\right)dy$
$\displaystyle \text{Putting } 1+y^{2}=t$
$\displaystyle \Rightarrow 2y\,dy=dt$
$\displaystyle \Rightarrow y\,dy=\frac{1}{2}dt$
$\displaystyle \Rightarrow x=-\frac{1}{2}\int\frac{1}{t}\,dt$
$\displaystyle \Rightarrow x=-\frac{1}{2}\log|t|+C$
$\displaystyle \Rightarrow x=-\frac{1}{2}\log|1+y^{2}|+C$
$\displaystyle \Rightarrow x+\frac{1}{2}\log|1+y^{2}|=C$
$\displaystyle \text{Hence, } x+\frac{1}{2}\log|1+y^{2}|=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 2: }~\frac{dy}{dx}=\frac{1+y^{2}}{y^{3}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{dy}{dx}=\frac{1+y^{2}}{y^{3}}$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{y^{3}}{1+y^{2}}$
$\displaystyle \Rightarrow dx=\frac{y^{3}}{1+y^{2}}\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \Rightarrow \int dx=\int\frac{y^{3}}{1+y^{2}}\,dy$
$\displaystyle \Rightarrow x=\int\frac{y^{3}}{1+y^{2}}\,dy$
$\displaystyle \Rightarrow x=\int\frac{y(1+y^{2})-y}{1+y^{2}}\,dy$
$\displaystyle \Rightarrow x=\int\left(y-\frac{y}{1+y^{2}}\right)dy$
$\displaystyle \Rightarrow x=\int y\,dy-\int\frac{y}{1+y^{2}}\,dy$
$\displaystyle \Rightarrow x=\frac{y^{2}}{2}-\int\frac{y}{1+y^{2}}\,dy$
$\displaystyle \text{Putting } 1+y^{2}=t$
$\displaystyle \Rightarrow 2y\,dy=dt$
$\displaystyle \Rightarrow y\,dy=\frac{1}{2}dt$
$\displaystyle \Rightarrow x=\frac{y^{2}}{2}-\frac{1}{2}\int\frac{1}{t}\,dt$
$\displaystyle \Rightarrow x=\frac{y^{2}}{2}-\frac{1}{2}\log|t|+C$
$\displaystyle \Rightarrow x=\frac{y^{2}}{2}-\frac{1}{2}\log|1+y^{2}|+C$
$\displaystyle \text{Hence, } x=\frac{y^{2}}{2}-\frac{1}{2}\log|1+y^{2}|+C \text{ is the required solution.}$

$\displaystyle \textbf{Question 3: }~\frac{dy}{dx}=\sin^{2}y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{dy}{dx}=\sin^{2}y$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{1}{\sin^{2}y}$
$\displaystyle \Rightarrow dx=\mathrm{cosec}^{2}y\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int dx=\int \mathrm{cosec}^{2}y\,dy$
$\displaystyle \Rightarrow x=-\cot y+C$
$\displaystyle \Rightarrow x+\cot y=C$
$\displaystyle \text{Hence, } x+\cot y=C \text{ is the required solution.}$

$\displaystyle \textbf{Question 4: }~\frac{dy}{dx}=\frac{1-\cos 2y}{1+\cos 2y}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{dy}{dx}=\frac{1-\cos 2y}{1+\cos 2y}$
$\displaystyle \Rightarrow \frac{dx}{dy}=\frac{1+\cos 2y}{1-\cos 2y}$
$\displaystyle \Rightarrow dx=\frac{1+\cos 2y}{1-\cos 2y}\,dy$
$\displaystyle \Rightarrow dx=\frac{2\cos^{2}y}{2\sin^{2}y}\,dy$
$\displaystyle \Rightarrow dx=\cot^{2}y\,dy$
$\displaystyle \text{Integrating both sides, we get}$
$\displaystyle \int dx=\int \cot^{2}y\,dy$
$\displaystyle \Rightarrow x=\int(\mathrm{cosec}^{2}y-1)\,dy$
$\displaystyle \Rightarrow x=\int \mathrm{cosec}^{2}y\,dy-\int dy$
$\displaystyle \Rightarrow x=-\cot y-y+C$
$\displaystyle \Rightarrow x+\cot y+y=C$
$\displaystyle \text{Hence, } x+\cot y+y=C \text{ is the required solution.}$