Question

Solve the each of the following differential equations:
(i) $\left(x-y\right)\frac{dy}{dx}=x+2y$

(ii) $x\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos \left(\frac{y}{x}\right)+x$

(iii) y dx + x log $\left(\frac{y}{x}\right)$ dy − 2x dy = 0

(iv) $\frac{dy}{dx}-y=\cos x$

(v) $x\frac{dy}{dx}+2y=x^2,x\ne 0$

(vi) $\frac{dy}{dx}+2y=\sin x$

(vii) $\frac{dy}{dx}+3y=e^{-2x}$

(viii) $\frac{dy}{dx}+\frac{y}{x}=x^2$

(ix) $\frac{dy}{dx}+\left(\sec x\right)y=\tan x$

(x) $x\frac{dy}{dx}+2y=x^2\log x$

(xi) $x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$

(xii) (1 + x²) dy + 2xy dx = cot x dx

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

(xiv) y dx + (x − y²) dy = 0

(xv) $\left(x+3y^2\right)\frac{dy}{dx}=y$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ \left(x-y\right)\frac{dy}{dx}=x+2y \\ \Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}.....\left(1\right) \\ \mathrm{Clearly}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{equation}, \\ \mathrm{Putting}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Substituting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx}\left(1\right)\mathrm{becomes}, \\ v+x\frac{dv}{dx}=\frac{x+2vx}{x-vx} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{1+2v}{1-v} \\ \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{1-v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^2}{1-v} \\ \Rightarrow x\frac{dv}{dx}=\frac{v^2+v+1}{1-v} \\ \Rightarrow \frac{1-v}{v^2+v+1}dv=\frac{1}{x}dx \\ \Rightarrow \left[\frac{-v}{v^2+v+1}+\frac{1}{v^2+v+1}\right]dv=\frac{1}{x}dx \\ \Rightarrow \left[-\frac{1}{2}\times \frac{2v+1-1}{v^2+v+1}+\frac{1}{v^2+v+1}\right]dv=\frac{1}{x}dx \\ \Rightarrow \left[-\frac{1}{2}\times \frac{2v+1}{v^2+v+1}+\frac{1}{2}\times \frac{1}{v^2+v+1}+\frac{1}{v^2+v+1}\right]dv=\frac{1}{x}dx \\ \Rightarrow \left[-\frac{1}{2}\times \frac{2v+1}{v^2+v+1}+\frac{3}{2}\times \frac{1}{v^2+v+1}\right]dv=\frac{1}{x}dx \\ \Rightarrow \left[-\frac{1}{2}\times \frac{2v+1}{v^2+v+1}+\frac{3}{2}\times \frac{1}{v^2+v+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}}}\right]dv=\frac{1}{x}dx \\ \Rightarrow \left[-\frac{1}{2}\times \frac{2v+1}{v^2+v+1}+\frac{3}{2}\times \frac{1}{{\left({\displaystyle v+\frac{1}{2}}\right)}^2+{\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2}}\right]dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \int \left[-\frac{1}{2}\times \frac{2v+1}{v^2+v+1}+\frac{3}{2}\times \frac{1}{{\left({\displaystyle v+\frac{1}{2}}\right)}^2+{\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2}}\right]dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{2}\int \frac{2v+1}{v^2+v+1}dv+\frac{3}{2}\int \frac{1}{{\left({\displaystyle v+\frac{1}{2}}\right)}^2+{\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2}}dv=\int \frac{1}{x}dx \\ \Rightarrow -\frac{1}{2}\log \left|v^2+v+1\right|+\frac{3}{2}\times \frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}}{\tan }^{-1}\frac{v+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\log \left|x\right|+C \\ \Rightarrow -\frac{1}{2}\log \left|{\left(\frac{y}{x}\right)}^2+\frac{y}{x}+1\right|+\frac{3}{2}\times \frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}}{\tan }^{-1}\frac{{\displaystyle \frac{y}{x}}+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\log \left|x\right|+C \\ \Rightarrow -\frac{1}{2}\log \left|\frac{y^2+xy+x^2}{x^2}\right|+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log \left|x\right|+C \\ \Rightarrow -\frac{1}{2}\log \left|y^2+xy+x^2\right|+\frac{1}{2}\log \left|x^2\right|+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log \left|x\right|+C \\ \Rightarrow -\frac{1}{2}\log \left|y^2+xy+x^2\right|+\log \left|x\right|+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log \left|x\right|+C \\ \Rightarrow -\frac{1}{2}\log \left|y^2+xy+x^2\right|+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=C \\ \Rightarrow \log \left|y^2+xy+x^2\right|-2\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=-2C \\ \Rightarrow \log \left|y^2+xy+x^2\right|=2\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)-2C \\ \Rightarrow \log \left|y^2+xy+x^2\right|=2\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)+k\mathrm{Where},k=-2C \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ x\cos \left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos \left(\frac{y}{x}\right)+x \\ \Rightarrow \frac{dy}{dx}=\frac{y\cos \left({\displaystyle \frac{y}{x}}\right)+x}{x\cos \left({\displaystyle \frac{y}{x}}\right)}.....\left(1\right) \\ \mathrm{Clearly}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{equation}, \\ \mathrm{Putting}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Substituting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx}\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=\frac{y\cos \left({\displaystyle \frac{y}{x}}\right)+x}{x\cos \left({\displaystyle \frac{y}{x}}\right)} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{vx\cos \left({\displaystyle v}\right)+x}{x\cos \left({\displaystyle v}\right)} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{v\cos \left({\displaystyle v}\right)+1}{\cos \left({\displaystyle v}\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{v\cos \left({\displaystyle v}\right)+1}{\cos \left({\displaystyle v}\right)}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{v\cos \left({\displaystyle v}\right)+1-v\cos \left(v\right)}{\cos \left({\displaystyle v}\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{1}{\cos \left({\displaystyle v}\right)} \\ \Rightarrow \cos \left(v\right)dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \cos \left(v\right)dv=\int \frac{1}{x}dx \\ \Rightarrow \sin \left(v\right)=\log \left|x\right|+\log \left|C\right| \\ \Rightarrow \sin \left(\frac{y}{x}\right)=\log \left|Cx\right| \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0 \\ \Rightarrow x\log \left(\frac{y}{x}\right)dy-2xdy=-ydx \\ \Rightarrow \left[\log \left(\frac{y}{x}\right)-2\right]xdy=-ydx \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{\left[\log \left({\displaystyle \frac{y}{x}}\right)-2\right]x} \\ \Rightarrow \frac{dy}{dx}=\frac{{\displaystyle \frac{y}{x}}}{2-\log \left({\displaystyle \frac{y}{x}}\right)}.....\left(1\right) \\ \mathrm{Clearly}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{homogenous}\mathrm{equation}, \\ \mathrm{Putting}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Substituting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx}\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{{\displaystyle v}}{2-\log \left({\displaystyle v}\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{{\displaystyle v}}{2-\log \left({\displaystyle v}\right)}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{{\displaystyle v-2v+v\log \left(v\right)}}{2-\log \left({\displaystyle v}\right)} \\ \Rightarrow x\frac{dv}{dx}=\frac{{\displaystyle -v+v\log \left(v\right)}}{2-\log \left({\displaystyle v}\right)} \\ \Rightarrow \frac{2-\log \left(v\right)}{-v+v\log \left(v\right)}dv=\frac{1}{x}dx \\ \Rightarrow \frac{\log \left(v\right)-2}{v\log \left(v\right)-v}dv=-\frac{1}{x}dx \\ \Rightarrow \frac{\log \left(v\right)-1-1}{v\left[\log \left(v\right)-1\right]}dv=-\frac{1}{x}dx \\ \Rightarrow \frac{\log \left(v\right)-1}{v\left[\log \left(v\right)-1\right]}dv-\frac{1}{v\left[\log \left(v\right)-1\right]}dv=-\frac{1}{x}dx \\ \Rightarrow \frac{1}{v}dv-\frac{1}{v\left[\log \left(v\right)-1\right]}dv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get} \\ \int \frac{1}{v}dv-\int \frac{1}{v\left[\log \left(v\right)-1\right]}dv=-\int \frac{1}{x}dx \\ \Rightarrow \log \left|v\right|-I=-\log \left|x\right|-\log C.....\left(2\right) \\ \mathrm{Where}, \\ I=\int \frac{1}{v\left[\log \left|\left(v\right)\right|-1\right]}dv \\ \mathrm{Puting}\log v=t \\ \frac{1}{v}dv=dt \\ \therefore I=\int \frac{1}{t-1}dt \\ \Rightarrow I=\log \left|t-1\right| \\ \Rightarrow I=\log \left|\log \left|v\right|-1\right|.....\left(3\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{get} \\ \log \left|v\right|-\log \left|\log \left|v\right|-1\right|=-\log \left|x\right|-\log C \\ \Rightarrow \log \left|\frac{v}{\log \left|v\right|-1}\right|=-\log \left|Cx\right| \\ \Rightarrow \frac{v}{\log \left|v\right|-1}=\frac{1}{Cx} \\ \Rightarrow \log \left|v\right|-1=vCx \\ \Rightarrow \log \left|\frac{y}{x}\right|-1=Cy \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}-y=\cos x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=-1 \\ Q=\cos x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{-1\int dx}=e^{-x} \\ \mathrm{Solution}\mathrm{is}\mathrm{given}\mathrm{by}, \\ y\times \mathrm{I}.\mathrm{F}.=\int \cos x\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{e}^{-x}=\int e^{-x}\cos xdx+C \\ \Rightarrow y{e}^{-x}=I+C.....\left(1\right) \\ \mathrm{Where}, \\ I=\int \underset{\mathrm{II}}{e^{-x}}\underset{\mathrm{I}}{\cos x}dx.....\left(2\right) \\ \Rightarrow I=\cos x\int e^{-x}dx-\int \left[\frac{d}{dx}\left(\cos x\right)\int e^{-x}dx\right]dx \\ \Rightarrow I=-\cos x{e}^{-x}-\int \sin x{e}^{-x}dx \\ \Rightarrow I=-\cos x{e}^{-x}-\int \underset{\mathrm{I}}{\sin x}\underset{\mathrm{II}}{e^{-x}}dx \\ \Rightarrow I=-\cos x{e}^{-x}-\sin x\int e^{-x}dx+\int \left[\frac{d}{dx}\left(\sin x\right)\int e^{-x}dx\right]dx \\ \Rightarrow I=-\cos x{e}^{-x}+\sin x{e}^{-x}-\int \left[\cos x{e}^{-x}\right]dx \\ \Rightarrow I=-\cos x{e}^{-x}+\sin x{e}^{-x}-I\left[\mathrm{Using}\left(2\right)\right] \\ \Rightarrow 2I=-\cos x{e}^{-x}+\sin x{e}^{-x} \\ \Rightarrow I=\frac{1}{2}\left(-\cos x+\sin x\right)e^{-x}.....\left(3\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \therefore y{e}^{-x}=\left(\sin x-\cos x\right)e^{-x}+C \\ \Rightarrow y=\frac{1}{2}\left(\sin x-\cos x\right)+C{e}^x \end{gathered}$$


$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+2y=x^2 \\ \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{2}{x} \\ Q=x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{2\int \frac{1}{x}dx} \\ =e^{2\log \left|x\right|} \\ =x^2 \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{x}^2=\int x^{3}dx+C \\ \Rightarrow y{x}^2=\frac{x^4}{4}+C \\ \Rightarrow y=\frac{x^2}{4}+C{x}^{-2} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+2y=\sin x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=2 \\ Q=\sin x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{2\int dx}=e^{2x} \\ \mathrm{Solution}\mathrm{is}\mathrm{given}\mathrm{by}, \\ y\times \mathrm{I}.\mathrm{F}.=\int \sin x\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{e}^{2x}=I+C.....\left(1\right) \\ \mathrm{Where}, \\ I=\int \underset{\mathrm{II}}{e^{2x}}\underset{\mathrm{I}}{\sin x}dx.....\left(2\right) \\ \Rightarrow I=\sin x\int e^{2x}dx-\int \left[\frac{d}{dx}\left(\sin x\right)\int e^{2x}dx\right]dx \\ \Rightarrow I=\frac{\sin x{e}^{2x}}{2}-\frac{1}{2}\int \cos x{e}^{2x}dx \\ \Rightarrow I=\frac{\sin x{e}^{2x}}{2}-\frac{1}{2}\int \underset{\mathrm{I}}{\cos x}\underset{\mathrm{II}}{e^{2x}}dx \\ \Rightarrow I=\frac{\sin x{e}^{2x}}{2}-\frac{1}{2}\cos x\int e^{2x}dx+\frac{1}{2}\int \left[\frac{d}{dx}\left(\cos x\right)\int e^{2x}dx\right]dx \\ \Rightarrow I=\frac{\sin x{e}^{2x}}{2}-\frac{1}{4}\cos x{e}^{2x}-\frac{1}{4}\int \left[\sin x{e}^{2x}\right]dx \\ \Rightarrow I=\frac{\sin x{e}^{2x}}{2}-\frac{1}{4}\cos x{e}^{2x}-\frac{1}{4}I\left[\mathrm{Using}\left(2\right)\right] \\ \Rightarrow I+\frac{1}{4}I=\frac{1}{2}\sin x{e}^{2x}-\frac{1}{4}\cos x{e}^{2x} \\ \Rightarrow \frac{5}{4}I=\frac{1}{4}\left(2\sin x{e}^{2x}-\cos x{e}^{2x}\right) \\ \Rightarrow I=\frac{1}{5}\left(2\sin x-\cos x\right)e^{2x}.....\left(3\right) \\ \mathrm{Therefore}\mathrm{from}\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \therefore y{e}^{2x}=\frac{1}{5}\left(2\sin x-\cos x\right)e^{2x}+C \\ \Rightarrow y=\frac{1}{5}\left(2\sin x-\cos x\right)+C{e}^{-2x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+3y=e^{-2x} \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=3 \\ Q=e^{-2x} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{3\int dx} \\ =e^{3x} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{e}^{3x}=\int e^{3x}\times e^{-2x}dx+C \\ \Rightarrow y{e}^{3x}=e^x+C \\ \Rightarrow y=e^{-2x}+C{e}^{-3x} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\frac{y}{x}=x^2 \\ \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=x^2 \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{x} \\ Q=x^2 \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{x}dx} \\ =e^{\log \left|x\right|} \\ =x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow yx=\int x^3+C \\ \Rightarrow xy=\frac{x^4}{4}+C \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ix}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\left(\sec x\right)y=\tan x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\sec x \\ Q=\tan x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \sec xdx} \\ =e^{\log \left|\left(\sec x+\tan x\right)\right|} \\ =\sec x+\tan x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y\left(\sec x+\tan x\right)=\int \left(\sec x+\tan x\right)\tan x+C \\ \Rightarrow y\left(\sec x+\tan x\right)=\int \sec x\times \tan xdx+\int {\tan }^{2}xdx+C \\ \Rightarrow y\left(\sec x+\tan x\right)=\int \sec x\times \tan xdx+\int \left({\sec }^{2}x-1\right)dx+C \\ \Rightarrow y\left(\sec x+\tan x\right)=\sec x+\tan x-x+C \end{gathered}$$


$$\begin{gathered} \left(\mathrm{x}\right)\mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+2y=x^2\log x \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}x,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}+\frac{2y}{x}=x\log x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{2}{x} \\ Q=x\log x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \frac{2}{x}dx} \\ =e^{2\log \left|x\right|} \\ =x^2 \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow x^{2}y=\int \underset{\mathrm{II}}{x^3}\underset{\mathrm{I}}{\log x}dx+C \\ \Rightarrow x^{2}y=\log x\int x^{3}dx-\int \left[\frac{d}{dx}\left(\log x\right)\int x^{3}dx\right]dx+C \\ \Rightarrow x^{2}y=\frac{x^4\log x}{4}-\int \frac{x^3}{4}dx+C \\ \Rightarrow x^{2}y=\frac{x^4\log x}{4}-\frac{x^4}{16}+C \\ \Rightarrow y=\frac{x^2\log x}{4}-\frac{x^2}{16}+\frac{C}{x^2} \\ \Rightarrow y=\frac{x^2}{16}\left(4\log x-1\right)+C{x}^{-2} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xi}\right)\mathrm{We}\mathrm{have}, \\ x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}x\mathit{}\log x,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x}\frac{\log x}{x\log x} \\ \Rightarrow \frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x^2} \\ \Rightarrow \frac{dy}{dx}+\left(\frac{1}{x\log x}\right)y=\frac{2}{x^2} \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{x\log x} \\ Q=\frac{2}{x^2} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{\int \frac{1}{x\log x}dx} \\ =e^{\log \left|\left(\log x\right)\right|} \\ =\log x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y\log x=2\int \frac{1}{x^2}\times \log xdx+C \\ \Rightarrow y\log x=I+C.....\left(1\right) \\ \mathrm{Where}, \\ I=2\int \underset{\mathrm{II}}{\frac{1}{x^2}}\underset{\mathrm{I}}{\log x}dx \\ \Rightarrow I=2\log x\int \frac{1}{x^2}dx-2\int \left[\frac{d}{dx}\left(\log x\right)\int \frac{1}{x^2}dx\right]dx \\ \Rightarrow I=\frac{-2}{x}\log x+2\int \left[\frac{1}{x^2}\right]dx \\ \Rightarrow I=\frac{-2}{x}\log x-\frac{2}{x}.....\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get} \\ \therefore y\log x=\frac{-2}{x}\log x-\frac{2}{x}+C \\ \Rightarrow y\log x=\frac{-2}{x}\left(\log x+1\right)+C \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xii}\right)\mathrm{We}\mathrm{have}, \\ \left(1+x^2\right)dy+2xydx=\cot xdx \\ \Rightarrow \frac{dy}{dx}+\frac{2x}{\left(1+x^2\right)}y=\frac{\cot x}{\left(1+x^2\right)} \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{2x}{\left(1+x^2\right)} \\ Q=\frac{\cot x}{\left(1+x^2\right)} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{2x}{\left(1+x^2\right)}dx} \\ =e^{\log \left|1+x^2\right|} \\ =1+x^2 \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y\left(1+x^2\right)=\int \left[\frac{\cot x}{\left(1+x^2\right)}\times \left(1+x^2\right)\right]dx+C \\ \Rightarrow y\left(1+x^2\right)=\int \cot xdx+C \\ \Rightarrow y\left(1+x^2\right)=\log \left|\sin x\right|+C \\ \Rightarrow y={\left(1+x^2\right)}^{-1}\log \sin x+C{\left(1+x^2\right)}^{-1} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xiii}\right)\mathrm{We}\mathrm{have}, \\ \left(x+y\right)\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\left(x+y\right)} \\ \mathrm{Let}x+y=v \\ \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1 \\ \therefore \frac{dv}{dx}-1=\frac{1}{v} \\ \Rightarrow \frac{dv}{dx}=\frac{1}{v}+1 \\ \Rightarrow \frac{dv}{dx}=\frac{1+v}{v} \\ \Rightarrow \frac{v}{1+v}dv=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{v}{1+v}dv=\int dx \\ \Rightarrow \int \frac{v+1-1}{1+v}dv=\int dx \\ \Rightarrow \int dv-\int \frac{1}{1+v}dv=\int dx \\ \Rightarrow v-\log \left|v+1\right|=x-\log C \\ \Rightarrow x+y-\log \left|x+y+1\right|=x-\log C \\ \Rightarrow y-\log \left|x+y+1\right|=-\log C \\ \Rightarrow y=\log \left|x+y+1\right|-\log C \\ \Rightarrow y=\log \left|\frac{x+y+1}{C}\right| \\ \Rightarrow C{e}^y=x+y+1 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xiv}\right)\mathrm{We}\mathrm{have}, \\ ydx+\left(x-y^2\right)dy=0 \\ \Rightarrow ydx=-\left(x-y^2\right)dy \\ \Rightarrow \frac{dx}{dy}=-\frac{1}{y}\left(x-y^2\right) \\ \Rightarrow \frac{dx}{dy}+\frac{1}{y}x=y.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where}P=\frac{1}{y}\mathrm{and}Q=y \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{\int \frac{1}{y}dy} \\ =e^{\log y}=y \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=y,\mathrm{we}\mathrm{get} \\ y\left(\frac{dx}{dy}+\frac{1}{y}x\right)=y\times y \\ \Rightarrow y\frac{dx}{dy}+x=y^2 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ xy=\int y^{2}dy+C \\ \Rightarrow xy=\frac{y^3}{3}+C \\ \Rightarrow x=\frac{y^2}{3}+\frac{C}{y} \\ \mathrm{Hence},x=\frac{y^2}{3}+\frac{C}{y}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{xv}\right)\mathrm{We}\mathrm{have}, \\ \left(x+3y^2\right)\frac{dy}{dx}=y \\ \Rightarrow \frac{dx}{dy}=\frac{1}{y}\left(x+3y^2\right) \\ \Rightarrow \frac{dx}{dy}-\frac{1}{y}x=3y.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where}P=-\frac{1}{y}\mathrm{and}Q=3y \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{-\int \frac{1}{y}dy} \\ =e^{-\log \left|y\right|}=\frac{1}{y} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\frac{1}{y},\mathrm{we}\mathrm{get} \\ \frac{1}{y}\left(\frac{dx}{dy}-\frac{1}{y}x\right)=\frac{1}{y}\times 3y \\ \Rightarrow \frac{1}{y}\left(\frac{dx}{dy}-\frac{1}{y}x\right)=3 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ x\frac{1}{y}=\int 3dy+C \\ \Rightarrow x\frac{1}{y}=3y+C \\ \Rightarrow x=3y^2+Cy \\ \mathrm{Hence},x=3y^2+Cy\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$