Differential Equations

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$\begin{array}{l}\left(vii\right)Thegivendifferentialequationis\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0\\ \Rightarrow \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{4x^2}{1+x^2}\\ Since,itisalineardifferentialequationwhereP=\frac{2x}{1+x^2}andQ=\frac{4x^2}{1+x^2}\\ I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{1+x^2}dx}}=e^{log\left(1+x^2\right)}=1+x^2\\ \therefore Solutionisy\times I.F.={\displaystyle \int Q\times I.F}.dx+c\\ \Rightarrow y.\left(1+x^2\right)={\displaystyle \int \frac{4x^2}{1+x^2}\times \left(1+x^2\right)dx+c}\\ \Rightarrow y.\left(1+x^2\right)={\displaystyle \int 4x^{2}dx+c}\Rightarrow y.\left(1+x^2\right)=\frac{4}{3}{x}^3+c\\ \Rightarrow y=\frac{4}{3}\frac{x^3}{\left(1+x^2\right)}+c{\left(1+x^2\right)}^{-1}\\ Hence,thesolutionisy=\frac{4}{3}\frac{x^3}{\left(1+x^2\right)}+c{\left(1+x^2\right)}^{-1}\\ \left(viii\right)Thegivendifferentialequationisydx+\left(x+xy\right)dy=0\\ \Rightarrow \left(x+xy\right)dy=-ydx\Rightarrow x\left(1+y\right)dy=-ydx\\ \Rightarrow \frac{1+y}{y}dy=-\frac{1}{x}dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{1+y}{y}dy=-{\displaystyle \int \frac{1}{x}dx}}\\ \Rightarrow {\displaystyle \int \left(\frac{1}{y}+1\right)dy=-{\displaystyle \int \frac{1}{x}dx}}\\ \Rightarrow logy+y=-logx+logc\\ \Rightarrow logy+logx+log{e}^y=logc\\ \Rightarrow log\left(xy.e^y\right)=logc\\ \Rightarrow xy=c.e^{-y}\\ Hence,thesolutionisxy=c.e^{-y}\end{array}$

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Sol:

$\begin{array}{l}Familyofconcentriccircleswithcentre\left(1,2\right)andradius\text{'}r\text{'}is\\ {\left(x-1\right)}^2+{\left(y-2\right)}^2=r^2\\ Differentiatingbothsidesw.r.t.x,weget\\ 2\left(x-1\right)+2\left(y-2\right)\frac{dy}{dx}=0\\ \Rightarrow \left(x-1\right)+\left(y-2\right)\frac{dy}{dx}=0\\ Whichistherequiredequation.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givendifferentialequationis\\ \left(1+y^2\right)ta{n}^{-1}xdx+2y\left(1+x^2\right)dy=0\\ \Rightarrow 2y\left(1+x^2\right)dy=-\left(1+y^2\right).ta{n}^{-1}x.dx\\ \Rightarrow \frac{2y}{1+y^2}dy=-\frac{ta{n}^{-1}x}{1+x^2}.dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{2y}{1+y^2}dy=-{\displaystyle \int \frac{ta{n}^{-1}x}{1+x^2}.dx}}\\ \Rightarrow log\left|1+y^2\right|=-\frac{1}{2}{\left(ta{n}^{-1}x\right)}^2+c\\ \Rightarrow \frac{1}{2}{\left(ta{n}^{-1}x\right)}^2+log\left|1+y^2\right|=c\\ Hence,therequiredsolutionis\frac{1}{2}{\left(ta{n}^{-1}x\right)}^2+log\left|1+y^2\right|=c.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}GivenequationisA{x}^2+B{y}^2=1\\ Differentiatingw.r.t.x,weget\\ \Rightarrow 2A.x+2By\frac{dy}{dx}=0\\ \Rightarrow A.x+By\frac{dy}{dx}=0\\ \Rightarrow By.\frac{dy}{dx}=-Ax\Rightarrow \frac{y}{x}.\frac{dy}{dx}=-\frac{A}{B}\\ Differentiatingbothsidesagainw.r.t.x,wehave\\ \frac{y}{x}.\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\left(\frac{x.\frac{dy}{dx}-y.1}{x^2}\right)=0\\ \Rightarrow \frac{y{x}^2}{x}.\frac{d^{2}y}{d{x}^2}+x.{\left(\frac{dy}{dx}\right)}^2-y.\frac{dy}{dx}=0\\ \Rightarrow xy.\frac{d^{2}y}{d{x}^2}+x.{\left(\frac{dy}{dx}\right)}^2-y.\frac{dy}{dx}=0\\ \Rightarrow xy.y^{\text{'}\text{'}}+x.{\left(y^{\text{'}}\right)}^2-y.y^{\text{'}}=0\\ Hence,therequiredequationisxy.y^{\text{'}\text{'}}+x.{\left(y^{\text{'}}\right)}^2-y.y^{\text{'}}=0\end{array}$

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Sol:

$\begin{array}{l}Givendifferentialequationisdy=cosx\left(2-ycosecx\right)dx\\ \Rightarrow \frac{dy}{dx}=cosx\left(2-ycosecx\right)\\ \Rightarrow \frac{dy}{dx}=2cosx-ycosx.cosecx\\ \Rightarrow \frac{dy}{dx}=2cosx-ycotx\\ \Rightarrow \frac{dy}{dx}+ycotx=2cosx\\ \therefore P=cotxandQ=2cosx\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int cotx.dx}}=e^{logsinx}=sinx\\ \therefore Solutionoftheequationis\\ y\times I.F.={\displaystyle \int Q\times I.F.dx+c}\\ \Rightarrow y.sinx={\displaystyle \int 2cosx.sinxdx+c}\\ \Rightarrow y.sinx={\displaystyle \int sin2xdx+c}\Rightarrow y.sinx=-\frac{1}{2}cos2x+c\\ Putx=\frac{\pi }{2}andy=2,weget\\ 2.sin\frac{\pi }{2}=-\frac{1}{2}cos2.\frac{\pi }{2}+c\\ \Rightarrow 2\left(1\right)=-\frac{1}{2}cos\pi +c\Rightarrow 2=-\frac{1}{2}\left(-1\right)+c\\ \Rightarrow 2=\frac{1}{2}+c\Rightarrow c=2-\frac{1}{2}=\frac{3}{2}\\ Hence,therequiredequationisysinx=-\frac{1}{2}cos2x+\frac{3}{2}.\end{array}$

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Sol:

$\begin{array}{l}Givendifferentialequationis\\ 2\left(y+3\right)-xy.\frac{dy}{dx}=0\\ \Rightarrow xy.\frac{dy}{dx}=2y+6\\ \Rightarrow \left(\frac{y}{2y+6}\right)dy=\frac{dx}{x}\\ \Rightarrow \frac{1}{2}\left(\frac{y}{y+3}\right)dy=\frac{dx}{x}\\ Integratingbothsidesweget\\ \frac{1}{2}{\displaystyle \int \frac{y}{y+3}.dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}{\displaystyle \int \left(1-\frac{3}{y+3}\right)dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}{\displaystyle \int 1.dy-\frac{3}{2}}{\displaystyle \int \frac{1}{y+3}dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}y-\frac{3}{2}log\left|y+3\right|=logx+c\\ Putx=1,y=-2\\ \Rightarrow \frac{1}{2}\left(-2\right)-\frac{3}{2}log\left|-2+3\right|=log\left(1\right)+c\\ \Rightarrow -1-\frac{3}{2}log\left(1\right)=log\left(1\right)+c\\ \Rightarrow -1-0=0+c\left[?log\left(1\right)=0\right]\\ \therefore c=-1\\ \therefore equationis\\ \frac{1}{2}y-\frac{3}{2}log\left|y+3\right|=logx-1\\ \Rightarrow y-3log\left|y+3\right|=2logx-2\\ \Rightarrow y-log\left|{\left(y+3\right)}^3\right|=log{x}^2-2\\ \Rightarrow log\left|{\left(y+3\right)}^3\right|+log{x}^2=y+2\\ \Rightarrow log\left|x^2{\left(y+3\right)}^3\right|=y+2\\ \Rightarrow x^2{\left(y+3\right)}^3=e^{y+2}\\ Hence,therequiredsol\end{array}$

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Sol:

$\begin{array}{l}Givendifferentialequationis\\ \left(x+y\right)\left(dx-dy\right)=dx+dy\\ \Rightarrow \left(x+y\right)dx-\left(x+y\right)dy=dx+dy\\ \Rightarrow -\left(x+y\right)dy-dy=dx-\left(x+y\right)dx\\ \Rightarrow -\left(x+y+1\right)dy=-\left(x+y-1\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{x+y-1}{x+y+1}\\ Putx+y=z\\ \therefore 1+\frac{dy}{dx}=\frac{dz}{dx}\\ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1\\ So,\frac{dz}{dx}-1=\frac{z-1}{z+1}\\ \Rightarrow \frac{dz}{dx}=\frac{z-1}{z+1}\Rightarrow \frac{dz}{dx}=\frac{z-1+z+1}{z+1}\\ \Rightarrow \frac{dz}{dx}=\frac{2z}{z+1}\Rightarrow \frac{z+1}{z}dz=2.dx\\ Integratingbothsidesweget\\ {\displaystyle \int \frac{z+1}{z}dz}=2{\displaystyle \int dx}\\ \Rightarrow {\displaystyle \int \left(1+\frac{1}{z}\right)dz}=2{\displaystyle \int dx}\\ \Rightarrow z+log\left|z\right|=2x+log\left|c\right|\\ \Rightarrow x+y+log\left|x+y\right|=2x+log\left|c\right|\\ \Rightarrow y+log\left|x+y\right|=x+log\left|c\right|\\ \Rightarrow log\left|x+y\right|=x-y+log\left|c\right|\\ \Rightarrow log\left|x+y\right|-log\left|c\right|=\left(x-y\right)\\ \Rightarrow log\left|\frac{x+y}{c}\right|=\left(x-y\right)\\ \Rightarrow \frac{x+y}{c}=e^{x-y}\\ \therefore x+y=c.e^{x-y}\\ Hence,therequiredsolutionisx+y=c.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationis{y}^{2}dx+\left(x^2-xy+y^2\right)dy=0\\ \Rightarrow \frac{dx}{dy}=-\frac{x^2-xy+y^2}{y^2}\\ Putx=vy\left[?Itisa\text{\hspace{0.17em}homogeneous}differentialequation\right]\\ \therefore \frac{dx}{dy}=v+y.\frac{dv}{dy}\\ \therefore v+y.\frac{dv}{dy}=-\left(\frac{y^2-v{y}^2+v^2{y}^2}{y^2}\right)\\ \Rightarrow v+y.\frac{dv}{dy}=-\frac{y^2\left(1-v+v^2\right)}{y^2}\\ \Rightarrow v+y.\frac{dv}{dy}=\left(-1+v-v^2\right)\\ \Rightarrow y.\frac{dv}{dy}=-1+v-v^2-v\\ \Rightarrow y.\frac{dv}{dy}=-1-v^2\Rightarrow \frac{dv}{v^2+1}=-\frac{dy}{y}\\ Integratingbothsidesweget\\ {\displaystyle \int \frac{dv}{v^2+1}}=-{\displaystyle \int \frac{dy}{y}}\\ \Rightarrow ta{n}^{-1}v=-logy+c\Rightarrow ta{n}^{-1}\left(\frac{x}{y}\right)+logy=c\\ Hence,therequiredsolutionista{n}^{-1}\left(\frac{x}{y}\right)+logy=c.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationis\left(1+y^2\right)+\left(x-e^{ta{n}^{-1}y}\right)\frac{dy}{dx}=0\\ \Rightarrow \left(x-e^{ta{n}^{-1}y}\right)\frac{dy}{dx}=-\left(1+y^2\right)\Rightarrow \frac{dy}{dx}=\frac{-\left(1+y^2\right)}{x-e^{ta{n}^{-1}y}}\\ \Rightarrow \frac{dx}{dy}=\frac{x-e^{ta{n}^{-1}y}}{-\left(1+y^2\right)}\Rightarrow \frac{dx}{dy}=-\frac{x}{\left(1+y^2\right)}+\frac{e^{ta{n}^{-1}y}}{1+y^2}\\ \Rightarrow \frac{dx}{dy}+\frac{x}{\left(1+y^2\right)}=\frac{e^{ta{n}^{-1}y}}{1+y^2}\\ \therefore P=\frac{1}{1+y^2}andQ=\frac{e^{ta{n}^{-1}y}}{1+y^2}\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int \frac{1}{1+y^2}.dy}}=e^{ta{n}^{-1}y}\\ \therefore Solutionoftheequationis\\ x\times I.F.={\displaystyle \int Q\times I.F.dy+c}\\ \Rightarrow y.e^{ta{n}^{-1}y}={\displaystyle \int \frac{e^{ta{n}^{-1}y}}{1+y^2}.e^{ta{n}^{-1}y}dy+c}\\ Put{e}^{ta{n}^{-1}y}=t\\ \therefore e^{ta{n}^{-1}y}.\frac{1}{1+y^2}dy=dt\\ \therefore x.e^{ta{n}^{-1}y}={\displaystyle \int t.d}t+c\\ \Rightarrow x.e^{ta{n}^{-1}y}=\frac{1}{2}{t}^2+c\\ \Rightarrow x.e^{ta{n}^{-1}y}=\frac{1}{2}{\left(e^{ta{n}^{-1}y}\right)}^2+c\Rightarrow x=\frac{1}{2}\left(e^{ta{n}^{-1}y}\right)+\frac{c}{e^{ta{n}^{-1}y}}\\ \Rightarrow 2x=e^{ta{n}^{-1}y}+\frac{2c}{e^{ta{n}^{-1}y}}\\ \Rightarrow 2x.e^{ta{n}^{-1}y}={\left(e^{ta{n}^{-1}y}\right)}^2+2c\\ Hence,therequiredsolutionis2x.e^{ta{n}^{-1}y}={\left(e^{ta{n}^{-1}y}\right)}^2+2c.\end{array}$