Maths NCERT Exemplar Solutions Class 12th Chapter Nine

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

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

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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:

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

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

Sol:

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

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

Sol:

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