Question

Solve each of the following initial value problems:
(i) $y\text{'}+y=e^x,y\left(0\right)=\frac{1}{2}$

(ii) $x\frac{dy}{dx}-y=\log x,y\left(1\right)=0$

(iii) $\frac{dy}{dx}+2y=e^{-2x}\sin x,y\left(0\right)=0$

(iv) $x\frac{dy}{dx}-y=\left(x+1\right)e^{-x},y\left(1\right)=0$

(v) $\left(1+y^2\right)dx+\left(x-e^{-{\tan }^{-1}y}\right)dx=0,y\left(0\right)=0$

(vi) $\frac{dy}{dx}+y\tan x=2x+x^2\tan x,y\left(0\right)=1$

(vii) $x\frac{dy}{dx}+y=x\cos x+\sin x,y\left(\frac{\mathrm{\pi }}{2}\right)=1$

(viii) $\frac{dy}{dx}+y\cot x=4x\mathrm{cosec}x,y\left(\frac{\mathrm{\pi }}{2}\right)=0$

(ix) $\frac{dy}{dx}+2y\tan x=\sin x;y=0\mathrm{when}x=\frac{\mathrm{\pi }}{3}$

(x) $\frac{dy}{dx}-3y\cot x=\sin 2x;y=2\mathrm{when}x=\frac{\mathrm{\pi }}{2}$
(xi)
(xii)

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ y\text{'}+y=e^x \\ \Rightarrow \frac{dy}{dx}+y=e^x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=1\mathrm{and}Q=e^x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int 1dx} \\ =e^x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^x,\mathrm{we}\mathrm{get} \\ {e}^x\left(\frac{dy}{dx}+y\right)=e^x{e}^x \\ \Rightarrow e^x\frac{dy}{dx}+e^{x}y=e^{2x} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^x=\int e^{2x}dx+C \\ \Rightarrow y{e}^x=\frac{e^{2x}}{2}+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(0\right)=\frac{1}{2} \\ \therefore \frac{1}{2}{e}^0=\frac{e^0}{2}+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y{e}^x=\frac{e^{2x}}{2} \\ \Rightarrow e^x=\frac{e^x}{2} \\ \mathrm{Hence},y=\frac{e^x}{2}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}-y=\log x \\ \Rightarrow \frac{dy}{dx}-\frac{y}{x}=\frac{\log x}{x}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-\frac{1}{x}\mathrm{and}Q=\frac{\log x}{x} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int \frac{1}{x}dx} \\ =e^{-\log x} \\ =\frac{1}{x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\frac{1}{x},\mathrm{we}\mathrm{get} \\ \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}\times \frac{\log x}{x} \\ \Rightarrow \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=\frac{\log x}{x^2} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\frac{1}{x}=\int \underset{\mathrm{II}}{\frac{1}{x^2}}\underset{\mathrm{I}}{\times \log x}dx+C \\ \Rightarrow \frac{y}{x}=\log x\int \frac{1}{x^2}dx-\int \left[\frac{d}{dx}\left(\log x\right)\int \frac{1}{x^2}dx\right]dx+C \\ \Rightarrow \frac{y}{x}=-\frac{\log x}{x}+\int \frac{1}{x^2}dx+C \\ \Rightarrow \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+C \\ \Rightarrow y=-\log x-1+Cx.....\left(2\right) \\ \mathrm{Now}, \\ y\left(1\right)=0 \\ \therefore 0=-0-1+C\left(1\right) \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y=-\log x-1+x \\ \Rightarrow y=x-1-\log x \\ \mathrm{Hence},y=x-1-\log x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+2y=e^{-2x}\sin x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=2\mathrm{and}Q=e^{-2x}\sin x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int 2dx} \\ =e^{2x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{2x},\mathrm{we}\mathrm{get} \\ {e}^{2x}\left(\frac{dy}{dx}+2y\right)=e^{2x}{e}^{-2x}\sin x \\ \Rightarrow e^{2x}\left(\frac{dy}{dx}+2y\right)=\sin x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{e}^{2x}=\int \sin xdx+C \\ \Rightarrow y{e}^{2x}=-\cos x+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(0\right)=0 \\ \therefore 0\times e^0=-\cos 0+C \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y{e}^{2x}=-\cos x+1 \\ \Rightarrow y{e}^{2x}=1-\cos x \\ \mathrm{Hence},y{e}^{2x}=1-\cos x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}-y=\left(x+1\right)e^{-x} \\ \Rightarrow \frac{dy}{dx}-\frac{1}{x}y=\left(\frac{x+1}{x}\right)e^{-x}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-\frac{1}{x}\mathrm{and}Q=\frac{x+1}{x}{e}^{-x} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-\int \frac{1}{x}dx} \\ =e^{-\log x} \\ =\frac{1}{x} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\frac{1}{x},\mathrm{we}\mathrm{get} \\ \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\frac{1}{x}\left(\frac{x+1}{x}\right)e^{-x} \\ \Rightarrow \frac{1}{x}\left(\frac{dy}{dx}-\frac{1}{x}y\right)=\left(\frac{x+1}{x^2}\right)e^{-x} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \frac{1}{x}y=\int \left(\frac{1}{x}+\frac{1}{x^2}\right)e^{-x}dx+C \\ \mathrm{Putting}\frac{1}{x}{e}^{-x}=t \\ \Rightarrow \left(-\frac{1}{x}{e}^{-x}-\frac{1}{x^2}{e}^{-x}\right)dx=dt \\ \Rightarrow \left(\frac{1}{x}+\frac{1}{x^2}\right)e^{-x}dx=-dt \\ \therefore \frac{1}{x}y=\int -dt+C \\ \Rightarrow \frac{y}{x}=-t+C \\ \Rightarrow \frac{y}{x}=-\frac{e^{-x}}{x}+C \\ \Rightarrow y=-e^{-x}+Cx.....\left(2\right) \\ \mathrm{Now}, \\ y\left(1\right)=0 \\ \therefore 0=-e^{-1}+C \\ \Rightarrow C=e^{-1} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y=-e^{-x}+x{e}^{-1} \\ \Rightarrow y=x{e}^{-1}-e^{-x} \\ \mathrm{Hence},y=x{e}^{-1}-e^{-x}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{We}\mathrm{have}, \\ \left(1+y^2\right)dx+\left(x-e^{-{\tan }^{-1}y}\right)dy=0 \\ \Rightarrow \left(x-e^{-{\tan }^{-1}y}\right)\frac{dy}{dx}=-\left(1+y^2\right) \\ \Rightarrow \left(1+y^2\right)\frac{dx}{dy}=-\left(x-e^{-{\tan }^{-1}y}\right) \\ \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{-{\tan }^{-1}y}}{1+y^2}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where}P=\frac{1}{1+y^2}\mathrm{and}Q=\frac{e^{-{\tan }^{-1}y}}{1+y^2} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{\int \frac{1}{1+y^2}dy} \\ =e^{{\tan }^{-1}y} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=e^{{\tan }^{-1}y},\mathrm{we}\mathrm{get} \\ {e}^{{\tan }^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^2}\right)=e^{{\tan }^{-1}y}\frac{e^{-{\tan }^{-1}y}}{1+y^2} \\ \Rightarrow e^{{\tan }^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^2}\right)=\frac{1}{1+y^2} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ {e}^{{\tan }^{-1}y}x=\int \frac{1}{1+y^2}dy+C \\ \Rightarrow x{e}^{{\tan }^{-1}y}={\tan }^{-1}y+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(0\right)=0 \\ \therefore 0\times e^0=0+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ x{e}^{{\tan }^{-1}y}={\tan }^{-1}y+0 \\ \Rightarrow x{e}^{{\tan }^{-1}y}={\tan }^{-1}y \\ \mathrm{Hence},x{e}^{{\tan }^{-1}y}={\tan }^{-1}y\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+y\tan x=2x+x^2\tan x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=\tan x\mathrm{and}Q=x^2\cot x+2x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \tan xdx} \\ =e^{\log \left|\sec x\right|}=\sec x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\sec x,\mathrm{we}\mathrm{get} \\ \sec x\left(\frac{dy}{dx}+y\tan x\right)=\sec x\left(x^2\tan x+2x\right) \\ \Rightarrow \sec x\left(\frac{dy}{dx}+y\tan x\right)=x^2\tan x\sec x+2x\sec x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\sec x=\int x^2\tan x\sec xdx+2\int \underset{\mathrm{II}}{x}\underset{\mathrm{I}}{\sec x}dx+C \\ \Rightarrow y\sec x=\int x^2\tan x\sec xdx+2secx\int xdx-2\int \left[\frac{d}{dx}\left(secx\right)\int xdx\right]dx+C \\ \Rightarrow y\sec x=\int x^2\tan x\sec xdx+x^2\sec x-\int x^2\tan x\sec xdx+C \\ \Rightarrow y\sec x=x^2\sec x+C \\ \Rightarrow y=x^2+C\cos x.....\left(2\right) \\ \mathrm{Now}, \\ y\left(0\right)=1 \\ \therefore 1=0+C\cos 0 \\ \Rightarrow C=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y=x^2+\cos x \\ \mathrm{Hence},y=x^2+\cos x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vii}\right)\mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+y=x\cos x+\sin x \\ \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=\cos x+\frac{\sin x}{x}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=\tan x\mathrm{and}Q=x^2\cot x+2x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \frac{1}{x}dx} \\ =e^{\log x}=x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=x,\mathrm{we}\mathrm{get} \\ x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x\left(\cos x+\frac{\sin x}{x}\right) \\ \Rightarrow x\left(\frac{dy}{dx}+\frac{1}{x}y\right)=x\cos x+\sin x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ xy=\int x\cos xdx+\int \sin xdx+C \\ \Rightarrow xy=\left[x\sin x-\int 1\left(\sin x\right)dx\right]-\cos x+C \\ \Rightarrow xy=x\sin x+\cos x-\cos x+C \\ \Rightarrow xy=x\sin x+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(\frac{\mathrm{\pi }}{2}\right)=1 \\ \therefore 1\times \frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{2}\sin \frac{\mathrm{\pi }}{2}+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ xy=x\sin x \\ \Rightarrow y=\sin x \\ \mathrm{Hence},y=\sin x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+y\cot x=4x\mathrm{cosec}x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=\cot x\mathrm{and}Q=4x\mathrm{cosec}x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \cot xdx} \\ =e^{\log \left|\sin x\right|}=\sin x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\sin x,\mathrm{we}\mathrm{get} \\ \sin x\left(\frac{dy}{dx}+y\cot x\right)=\sin x\left(4x\mathrm{cosec}x\right) \\ \Rightarrow \sin x\left(\frac{dy}{dx}+y\cot x\right)=4x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\sin x=4\int xdx+C \\ \Rightarrow y\sin x=2x^2+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(\frac{\mathrm{\pi }}{2}\right)=0 \\ \therefore 0\times \sin \left(\frac{\mathrm{\pi }}{2}\right)=2{\left(\frac{\mathrm{\pi }}{2}\right)}^2+C \\ \Rightarrow C=-\frac{{\mathrm{\pi }}^2}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y\sin x=2x^2-\frac{{\mathrm{\pi }}^2}{2} \\ \mathrm{Hence},y\sin x=2x^2-\frac{{\mathrm{\pi }}^2}{2}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ix}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+2y\tan x=\sin x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=2\tan x\mathrm{and}Q=\sin x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{2\int \tan xdx} \\ =e^{2\log \left|\sec x\right|}={\sec }^{2}x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.={\sec }^{2}x,\mathrm{we}\mathrm{get} \\ {\sec }^{2}x\left(\frac{dy}{dx}+2y\tan x\right)={\sec }^{2}x\times \sin x \\ \Rightarrow {\sec }^{2}x\left(\frac{dy}{dx}+2y\tan x\right)=\tan x\sec x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{\sec }^{2}x=\int \tan x\sec xdx+C \\ \Rightarrow y{\sec }^{2}x=\sec x+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(\frac{\mathrm{\pi }}{3}\right)=0 \\ \therefore 0{\left(\sec \frac{\mathrm{\pi }}{3}\right)}^2=\sec \frac{\mathrm{\pi }}{3}+C \\ \Rightarrow C=-2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y{\sec }^{2}x=\sec x-2 \\ \Rightarrow y=\cos x-2{\cos }^{2}x \\ \mathrm{Hence},y=\cos x-2{\cos }^{2}x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{x}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-3y\cot x=\sin 2x.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=-3\cot x\mathrm{and}Q=\sin 2x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{-3\int \cot xdx} \\ =e^{-3\log \left|\sin x\right|}={\mathrm{cosec}}^{3}x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.={\mathrm{cosec}}^{3}x,\mathrm{we}\mathrm{get} \\ {\mathrm{cosec}}^{3}x\left(\frac{dy}{dx}-3y\cot x\right)=\sin 2x\left({\mathrm{cosec}}^{3}x\right) \\ \Rightarrow {\mathrm{cosec}}^{3}x\left(\frac{dy}{dx}-3y\cot x\right)=2\cot x\mathrm{cosec}x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y{\mathrm{cosec}}^{3}x=2\int \cot x\mathrm{cosec}xdx+C \\ \Rightarrow y{\mathrm{cosec}}^{3}x=-2\mathrm{cosec}x+C \\ \Rightarrow y=-2{\sin }^{2}x+C{\sin }^{3}x.....\left(2\right) \\ \mathrm{Now}, \\ y\left(\frac{\mathrm{\pi }}{2}\right)=2 \\ \therefore 2=-2{\sin }^2\frac{\mathrm{\pi }}{2}+C{\sin }^3\frac{\mathrm{\pi }}{2} \\ \Rightarrow C=4 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y=-2{\sin }^{2}x+4{\sin }^{3}x \\ \Rightarrow y=4{\sin }^{3}x-2{\sin }^{2}x \\ \mathrm{Hence},y=4{\sin }^{3}x-2{\sin }^{2}x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xi}\right)\frac{dy}{dx}+y\cot x=2\cos x,y\left(\frac{\mathrm{\pi }}{2}\right)=0 \\ \frac{dy}{dx}+y\cot x=2\cos x....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=\cot x\mathrm{and}Q=2\cos x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \cot xdx} \\ =e^{\mathrm{logsin}x} \\ =\sin x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\sin x,\mathrm{we}\mathrm{get} \\ \sin x\left(\frac{dy}{dx}+y\cot x\right)=2\sin x\cos x \\ \Rightarrow \sin x\frac{dy}{dx}+y\cos x=\sin 2x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\sin x=\int \sin 2xdx+C \\ \Rightarrow y\sin x=-\frac{\cos 2x}{2}+C.....\left(2\right) \\ \mathrm{Now}, \\ y\left(\frac{\mathrm{\pi }}{2}\right)=0 \\ \therefore 0\times \sin \left(\frac{\mathrm{\pi }}{2}\right)=-\frac{\mathrm{cos\pi }}{2}+C \\ \Rightarrow C=-\frac{1}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ y\sin x=-\frac{\cos 2x}{2}-\frac{1}{2} \\ \Rightarrow 2y\sin x=-\left(1+\cos 2x\right) \\ \Rightarrow 2y\sin x=-2{\cos }^{2}x \\ \Rightarrow y=-\cot x\cos x \\ \mathrm{Hence},y=-\cot x\cos x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(xii\right)dy=\cos x\left(2-y\cos ecx\right)dx \\ \Rightarrow \frac{dy}{dx}=2\cos x-ycotx \\ \Rightarrow \frac{dy}{dx}+y\cot x=2\cos x....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where}P=\cot x\mathrm{and}Q=2\cos x \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \cot xdx} \\ =e^{\mathrm{logsin}x} \\ =\sin x \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\sin x,\mathrm{we}\mathrm{get} \\ \sin x\left(\frac{dy}{dx}+y\cot x\right)=2\sin x\cos x \\ \Rightarrow \sin x\frac{dy}{dx}+y\cos x=\sin 2x \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ y\sin x=\int \sin 2xdx+C \\ \Rightarrow y\sin x=-\frac{\cos 2x}{2}+C \\ \mathrm{Hence},y\sin x=-\frac{\cos 2x}{2}+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$