Class 12th

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

Sol:

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

Sol:

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

Sol:

$\begin{array}{l}Givenequationis\left(1+t\right)\frac{dy}{dt}-ty=1\\ \Rightarrow \frac{dy}{dt}-\left(\frac{t}{1+t}\right)y=\frac{1}{1+t}\\ \therefore P=-\frac{t}{1+t}andQ=\frac{1}{1+t}\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdt}}=e^{{\displaystyle \int -\frac{t}{1+t}.dt}}=e^{-{\displaystyle \int \frac{1+t-1}{1+t}.dt}}\\ =e^{-{\displaystyle \int \left(1-\frac{1}{1+t}\right).dt}}=e^{-\left[t-log\left(1+t\right)\right]}\\ =e^{-t+log\left(1+t\right)}=e^{-t}.e^{log\left(1+t\right)}=e^{-t}\left(1+t\right)\\ \therefore Solutionoftheequationis\\ y*I.F.={\displaystyle \int Q.I.F.dt+c}\\ \Rightarrow y.e^{-t}\left(1+t\right)={\displaystyle \int \frac{1}{\left(1+t\right)}.e^{-t}\left(1+t\right)dt+c}\\ \Rightarrow y.e^{-t}\left(1+t\right)={\displaystyle \int e^{-t}dt+c}\\ \Rightarrow y.e^{-t}\left(1+t\right)=-e^{-t}+c\\ Putt=0andy=-1\left[?y\left(0\right)=-1\right]\\ \Rightarrow -1.e^0.1=-e^0+c\\ \Rightarrow -1=-1+c\Rightarrow c=0\\ Sotheequationbecomes\\ \therefore y.e^{-t}\left(1+1\right)=-e^{-t}\\ \Rightarrow 2y=-1\Rightarrow y=-\frac{1}{2}\\ Hence,y\left(1\right)=-\frac{1}{2}isverified.\end{array}$

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

Sol:

$\begin{array}{l}Givenequationis\left(\frac{2+sinx}{1+y}\right)\frac{dy}{dx}=-cosx\\ \Rightarrow \left(\frac{2+sinx}{cosx}\right)\frac{dy}{dx}=-\left(1+y\right)\Rightarrow \frac{dy}{\left(1+y\right)}=-\left(\frac{cosx}{2+sinx}\right)dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dy}{\left(1+y\right)}=-{\displaystyle \int \left(\frac{cosx}{2+sinx}\right)dx}}\\ \Rightarrow log\left|1+y\right|=-log\left|2+sinx\right|+logc\\ \Rightarrow log\left|1+y\right|+log\left|2+sinx\right|=logc\\ \Rightarrow log\left(1+y\right)\left(2+sinx\right)=logc\\ \Rightarrow \left(1+y\right)\left(2+sinx\right)=c\\ Putx=0andy=1,weget\\ \Rightarrow \left(1+1\right)\left(2+sin0\right)=c\Rightarrow c=4\\ \therefore equationis\left(1+y\right)\left(2+sinx\right)=4\\ Nowputx=\frac{\pi }{2}\\ \therefore \left(1+y\right)\left(2+sin\frac{\pi }{2}\right)=4\Rightarrow \left(1+y\right)\left(2+1\right)=4\\ \Rightarrow 1+y=\frac{4}{3}\Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\\ So,y\left(\frac{\pi }{2}\right)=\frac{1}{3}\\ Hence,therequiredsolutionisy\left(\frac{\pi }{2}\right)=\frac{1}{3}.\end{array}$

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

Sol:

$\begin{array}{l}Givendifferentialequationis\left(x+2y^3\right)\frac{dy}{dx}=y\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x+2y^3}\Rightarrow \frac{dx}{dy}=\frac{x+2y^3}{y}\\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}+\frac{2y^3}{y}\Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y^2\\ \therefore P=-\frac{1}{y}andQ=2y^2\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{y}.dx}}=e^{-logy}=e^{log\frac{1}{y}}=\frac{1}{y}\\ \therefore Solutionoftheequationis\\ x*I.F.={\displaystyle \int Q.I.F.dy+c}\\ \Rightarrow x.\frac{1}{y}={\displaystyle \int 2y^2.\frac{1}{y}dy+c}\Rightarrow \frac{x}{y}=2{\displaystyle \int ydy+c}\\ \Rightarrow \frac{x}{y}=2.\frac{y^2}{2}+c\Rightarrow \frac{x}{y}=y^2+c\\ \therefore x=y^3+cy=y\left(y^2+c\right)\\ Hence,therequiredvalueisx=y\left(y^2+c\right).\end{array}$

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

Sol:

 $\begin{array}{l}Givenequationis\frac{dy}{dx}=1+x+y^2+x{y}^2\\ \Rightarrow \frac{dy}{dx}=1\left(1+x\right)+y^2\left(1+x\right)\\ \Rightarrow \frac{dy}{dx}=\left(1+x\right)\left(1+y^2\right)\Rightarrow \frac{dy}{1+y^2}\Rightarrow \left(1+x\right)dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dy}{1+y^2}}={\displaystyle \int \left(1+x\right)dx}\Rightarrow ta{n}^{-1}y=x+\frac{x^2}{2}+c\\ Putx=0andy=0,wegetta{n}^{-1}\left(0\right)=0+0+c\Rightarrow c=0\\ \therefore ta{n}^{-1}y=x+\frac{x^2}{2}y=tan\left(x+\frac{x^2}{2}\right)\\ Hence,therequiredsolutionisy=tan\left(x+\frac{x^2}{2}\right).\end{array}$

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

Sol:

$\begin{array}{l}Givenequationisydx-xdy=x^{2}ydx\\ \Rightarrow ydx-x^{2}ydx=xdy\\ \Rightarrow y\left(1-x^2\right)dx=xdy\\ \Rightarrow \left(\frac{1-x^2}{x}\right)dx=\frac{dy}{y}\Rightarrow \left(\frac{1}{x}-x\right)dx=\frac{dy}{y}\\ Integratingbothsides,wehave\\ {\displaystyle \int \left(\frac{1}{x}-x\right)dx}={\displaystyle \int \frac{dy}{y}}\Rightarrow logx-\frac{x^2}{2}=logy+logc\\ \Rightarrow logx-\frac{x^2}{2}=logyc\Rightarrow logx-logyc=\frac{x^2}{2}\\ \Rightarrow log\frac{x}{yc}=\frac{x^2}{2}\Rightarrow \frac{x}{yc}=e^{\frac{x^2}{2}}\Rightarrow \frac{yc}{x}=e^{\frac{-x^2}{2}}\\ \Rightarrow yc=x{e}^{\frac{-x^2}{2}}\Rightarrow y=\frac{1}{c}.x{e}^{\frac{-x^2}{2}}\Rightarrow y=kx{e}^{\frac{-x^2}{2}}\left[?k=\frac{1}{c}\right]\\ Hence,therequiredsolutioniskx{e}^{\frac{-x^2}{2}}.\end{array}$

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

Sol:

$\begin{array}{l}Givenequationis\frac{dy}{dx}+1=e^{x+y}\\ Putx+y=t\\ \therefore 1+\frac{dy}{dx}=\frac{dt}{dx}\\ \therefore \frac{dt}{dx}=e^t\Rightarrow \frac{dt}{e^t}=dx\Rightarrow e^{-t}dt=dx\\ Integratingbothsides,wehave\\ {\displaystyle \int e^{-t}dt}={\displaystyle \int dx}\Rightarrow -e^{-t}=x+c\\ -e^{-\left(x+y\right)}=x+c\Rightarrow \frac{-1}{e^{\left(x+y\right)}}x+c\Rightarrow \left(x+c\right).e^{\left(x+y\right)}=-1\\ Hence,therequiredsolutionis\left(x+c\right).e^{\left(x+y\right)}+1=0.\end{array}$