Class 12th

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

Sol:

$\begin{array}{l}Givendifferentialequationis\left(x^2-1\right)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}\\ Divindingby\left(x^2-1\right),weget\\ \frac{dy}{dx}+\frac{xy}{x^2-1}\Rightarrow \frac{1}{{\left(x^2-1\right)}^2}\\ Itisalineardifferentialequationoffirstorderandfirstdegree.\\ \therefore P=\frac{2x}{x^2-1}andQ=\frac{1}{{\left(x^2-1\right)}^2}\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{x^2-1}dx}}=e^{log\left(x^2-1\right)}=\left(x^2-1\right)\\ \therefore Solutionoftheequationis\\ y*I.F.={\displaystyle \int Q.I.F.dx+c}\\ \Rightarrow y\left(x^2-1\right)={\displaystyle \int \frac{1}{x^2-1}dx+c\Rightarrow y\left(x^2-1\right)=\frac{1}{2}log\left|\frac{x-1}{x+1}\right|}+c\\ Hence,therequiredvalueisy\left(x^2-1\right)=\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+c.\end{array}$

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

Sol:

$\begin{array}{l}Givenequationis\frac{dy}{dx}=e^{-2y}\\ \Rightarrow \frac{dy}{e^{-2y}}=dx\Rightarrow e^{2y}.dy=dx\\ Integratingbothsides,weget\\ {\displaystyle \int e^{2y}.dy={\displaystyle \int dx}}\Rightarrow \frac{1}{2}{e}^{2y}=x+c\\ Puty=0andx=5\\ \Rightarrow \frac{1}{2}{e}^{2y}=5+c\Rightarrow c=\frac{1}{2}-5=-\frac{9}{2}\\ \therefore Theequationbecomes\frac{1}{2}{e}^{2y}=x-\frac{9}{2}\\ Nowputtingy=3,weget\\ \frac{1}{2}{e}^6=x-\frac{9}{2}\Rightarrow x=\frac{1}{2}{e}^6+\frac{9}{2}\\ Hence,therequiredvalueofx=\frac{1}{2}\left(e^6+9\right).\end{array}$

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

Sol:

$\begin{array}{l}Equationofallnonverticallinesarey=mx+c\\ Differentiatingwithrespecttox,weget\frac{dy}{dx}=m\\ Againdifferentiatingw.r.t.xwehave\frac{d^{2}y}{d{x}^2}=0\\ Hence,therequiredequationis\frac{d^{2}y}{d{x}^2}=0.\end{array}$

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

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}=2^{y-x}\Rightarrow \frac{dy}{dx}=\frac{2^y}{2^x}\\ Separatingthevariables,weget\\ \frac{dy}{2^y}=\frac{dx}{2^x}\Rightarrow 2^{-y}dy=2^{-x}dx\\ Integratingbothsides,weget\\ {\displaystyle \int 2^{-y}dy}={\displaystyle \int 2^{-x}dx}\\ \frac{-2^{-y}}{log2}=\frac{-2^{-x}}{log2}+c\Rightarrow -2^{-y}=-2^{-x}+clog2\\ \Rightarrow -2^{-y}+2^{-x}=clog2\\ \Rightarrow 2^{-x}-2^{-y}=k\left[whereclog2=k\right]\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx\frac{dy}{dx}=y\left(logy-logx+1\right)\\ \Rightarrow x\frac{dy}{dx}=y\left[log\left(\frac{y}{x}\right)+1\right]\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left[log\left(\frac{y}{x}\right)+1\right]\\ Since,itisadifferentialequation\\ So,puty=vx\Rightarrow \frac{dy}{dx}=v+x.\frac{dv}{dx}\\ \Rightarrow v+x.\frac{dv}{dx}=\frac{vx}{x}\left[log\left(\frac{vx}{x}\right)+1\right]\\ \Rightarrow v+x.\frac{dv}{dx}=v\left[logv+1\right]\\ \Rightarrow x.\frac{dv}{dx}=v\left[logv+1\right]-v\Rightarrow x.\frac{dv}{dx}=v\left[logv+1-1\right]\\ \Rightarrow x.\frac{dv}{dx}=v.logv\Rightarrow \frac{dv}{vlogv}=\frac{dx}{x}\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dv}{vlogv}={\displaystyle \int \frac{dx}{x}}}\\ Putlogv=tonL.H.S.\\ \Rightarrow \frac{1}{v}dv=dt\\ \therefore {\displaystyle \int \frac{dt}{t}}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow log\left|t\right|=log\left|x\right|+logc\\ \Rightarrow log\left|logv\right|=logxc\Rightarrow logv=xc\\ \Rightarrow log\left(\frac{y}{x}\right)=xc\\ Hence,therequiredsolutionislog\left(\frac{y}{x}\right)=xc.\end{array}$

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

Sol:

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

Sol:

$\begin{array}{l}Here,slopeoftheofthecurve=\frac{dy}{dx}\\ andthedifferencebetweentheabscissaandordinate=x-y.\\ \therefore Asperthecondition,\frac{dy}{dx}={\left(x-y\right)}^2\\ Putx-y=v\\ 1-\frac{dy}{dx}=\frac{dv}{dx}\\ \therefore \frac{dy}{dx}=1-\frac{dv}{dx}\\ \therefore theequationbecomes\\ 1-\frac{dv}{dx}=v^2\Rightarrow \frac{dv}{dx}=1-v^2\Rightarrow \frac{dv}{1-v^2}=dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dv}{1-v^2}={\displaystyle \int dx}}\\ \Rightarrow \frac{1}{2}log\left|\frac{1+v}{1-v}\right|=x+c\Rightarrow \frac{1}{2}log\left|\frac{1+x-y}{1-x+y}\right|=x+c\dots \left(1\right)\\ Since,thecurveisthrough\left(0,0\right),then\\ \Rightarrow \frac{1}{2}log\left|\frac{1+0-0}{1-0+0}\right|=0+c\Rightarrow c=0\\ \therefore Onputtingc=0ineq.\left(1\right)weget\\ \frac{1}{2}log\left|\frac{1+x-y}{1-x+y}\right|=x\Rightarrow log\left|\frac{1+x-y}{1-x+y}\right|=2x\\ \therefore \frac{1+x-y}{1-x+y}=e^{2x}\\ \Rightarrow \left(1+x-y\right)=e^{2x}\left(1-x+y\right)\\ Hence,therequiredequationis\left(1+x-y\right)=e^{2x}\left(1-x+y\right).\end{array}$

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

Sol:

$\begin{array}{l}Giventhattheslopeoftangenttoacurveat\left(x,y\right)is\\ \frac{dy}{dx}=\frac{y-1}{x^2+x}\Rightarrow \frac{dy}{y-1}=\frac{dx}{x^2+x}\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x}}}\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x+\frac{1}{4}-\frac{1}{4}}}}\left[Makingperfectsquare\right]\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{{\left(x+\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}}\\ \Rightarrow log\left|y-1\right|=\frac{1}{2*\frac{1}{2}}log\left|\frac{x+\frac{1}{2}-\frac{1}{2}}{x+\frac{1}{2}+\frac{1}{2}}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|\frac{x}{x+1}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|c\left(\frac{x}{x+1}\right)\right|\\ \therefore y-1=\frac{cx}{x+1}\Rightarrow \left(y-1\right)\left(x+1\right)=cx\\ Since,thelineispassingthroughthepoint\left(1,0\right),then\\ \left(0-1\right)\left(1+1\right)=c\left(1\right)\Rightarrow c=2\\ Hence,therequiredsolutionis\left(y-1\right)\left(x+1\right)=2x.\end{array}$