Maths NCERT Exemplar Solutions Class 12th Chapter Nine

IF =e^x

$y{e}^x={\displaystyle \int \frac{e^x}{1+e^{2x}}dx}$

${\displaystyle \int \frac{dt}{1+t^2}}$

= tan^-1 (t) + c

$\underset{x\to \infty }{lim}{e}^{x}y=\underset{x\to \infty }{lim}ta{n}^{-1}{e}^x+\frac{\pi }{4}=\frac{3\pi }{4}$

 $\underset{x\to 0}{lim}\frac{\alpha e^x+\beta e^{-x}+\gamma sinx}{xsi{n}^{2}x}$

$=\underset{x\to 0}{lim}\frac{\alpha e^x+\beta e^{-x}+\gamma sinx}{x^3}$

$\Rightarrow \alpha +\beta =0,\alpha -\beta +\gamma =0,\frac{\alpha +\beta }{2}=0,\frac{\alpha }{6}-\frac{\beta }{6}-\frac{\gamma }{6}=\frac{2}{3}$

$\Rightarrow \beta =-\alpha \gamma =-2\alpha \alpha -\beta -\gamma =4\Rightarrow \alpha +\alpha +2\alpha =4\Rightarrow \gamma =1$

$\alpha =1,\beta =-1,\gamma =-2$

$x\frac{dy}{dx}-y=x^2(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x),x>0$

$\frac{dy}{dx}-\frac{y}{x}=x(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x)\Rightarrow \frac{dy}{dx}+Py=Q$

So, I.F. $=e^{\int -\frac{1}{x}dx}\frac{1}{\left|x\right|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}\left(x\right(x\mathrm{c}\mathrm{o}\mathrm{s}?x+\mathrm{s}\mathrm{i}\mathrm{n}?x\left)\right)dx$

$\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\mathrm{s}\mathrm{i}\mathrm{n}?\mathrm{x}+\mathrm{C}$

$?\mathrm{y}\left(\pi \right)=\pi \Rightarrow \mathrm{C}=1$

So, $y=x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+x\Rightarrow (y{)}_{\pi /2}=\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Also, $\frac{dy}{dx}=x^2\mathrm{c}\mathrm{o}\mathrm{s}?x+2x\mathrm{s}\mathrm{i}\mathrm{n}?x+1$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-x^2\mathrm{s}\mathrm{i}\mathrm{n}?x+4x\mathrm{c}\mathrm{o}\mathrm{s}?x+2\mathrm{s}\mathrm{i}\mathrm{n}?x$

$\mathrm{u}=\frac{2\mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{k}\mathrm{i}}$

$=\frac{2x^2+(2y+1)(y-k)}{x^2+(y-k{)}^2}+i\frac{\left(x\right(2y+1)-2x(y-k\left)\right)}{x^2+(y-k{)}^2}$

Since $\mathrm{R}\mathrm{e}?\left(u\right)+\mathrm{I}\mathrm{m}?\left(u\right)=1$

$\Rightarrow 2{\mathrm{x}}^2+(2\mathrm{y}+1)(\mathrm{y}-\mathrm{k})+\mathrm{x}(2\mathrm{y}+1)-2\mathrm{x}(\mathrm{y}-\mathrm{k})={\mathrm{x}}^2+(\mathrm{y}-\mathrm{k}{)}^2$

$\begin{array}{c}\mathrm{P}\left(0,{\mathrm{y}}_1\right)\\ \mathrm{Q}\left(0,{\mathrm{y}}_2\right)\end{array}?\Rightarrow {\mathrm{y}}^2+\mathrm{y}-\mathrm{k}-{\mathrm{k}}^2=0?\begin{array}{c}{\mathrm{y}}_1+{\mathrm{y}}_2=-1\\ {\mathrm{y}}_1{\mathrm{y}}_2=-\mathrm{k}-{\mathrm{k}}^2\end{array}$

$?\mathrm{P}\mathrm{Q}=5$

$\Rightarrow \left|{\mathrm{y}}_1-{\mathrm{y}}_2\right|=5\Rightarrow {\mathrm{k}}^2+\mathrm{k}-6=0$

$\Rightarrow \mathrm{k}=-\mathrm{3,2}$

So, $k=2(k>0)$

$\left(1-x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt[]{1-x^2}\right)dx$

$\therefore \frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{x^3+3}{\sqrt[]{1-x^2}}$

$\therefore l.F.=e^{{\displaystyle \int \frac{X}{1-x^2}dx}}=\sqrt[]{1-x^2}$

$\therefore y\left(x\right)=\frac{x^4+12x}{4\sqrt[]{1-x^2}}$

$\therefore {\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\sqrt[]{1-x^2}y\left(x\right)dx={\displaystyle \underset{\frac{-1}{2}}{\overset{\frac{1}{2}}{\int }}\left(\frac{x^4+12x}{4}\right)}dx}$

$\therefore k=\frac{1}{320}$

$\therefore k^{-1}=320$

$\left(ta{n}^{-1}y\right)-x)dy=\left(1+y^2\right)dx$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

$l.F=e^{{\displaystyle \int \frac{1}{1+y^2}dy}}=e^{ta{n}^{-1}y}$

$x.e^{ta{n}^{-1}y}{\displaystyle \int \frac{e^{ta{n}^{-1}y}ta{n}^{-1}y}{1+y^2}dy}$

Let

$e^{ta{n}^{-1}y}=t$

$=x{e}^{ta{n}^{-1}y}=e^{ta{n}^{-1}y}y-e^{ta{n}^{-1}y}+c...(i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x=\frac{2}{6}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{{\left(1+x^2\right)}^2}\\ \text{since\hspace{0.17em}}itislineardifferentialequationwhereP=\frac{2xy}{1+x^2}andQ=\frac{1}{{\left(1+x^2\right)}^2}\\ \therefore I.F.=e^{{\displaystyle \int P.dx}}=e^{{\displaystyle \int \frac{2xy}{1+x^2}.dx}}=e^{log\left(1+x^2\right)}=\left(1+x^2\right)\\ So,thesolutionis\\ y*I.F.={\displaystyle \int Q*}I.F.dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^2}*\left(1+x^2\right)}dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{1}{\left(1+x^2\right)}.}dx+c\\ \Rightarrow y\left(1+x^2\right)=ta{n}^{-1}x+c\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}=e^{x-y}+x^2{e}^{-y}\\ \Rightarrow \frac{dy}{dx}=e^x.e^{-y}+x^2{e}^{-y}\Rightarrow \frac{dy}{dx}=e^{-y}\left(e^x+x^2\right)\\ \Rightarrow \frac{dy}{e^{-y}}=\left(e^x+x^2\right)dx\Rightarrow e^y.dy=\left(e^x+x^2\right)dx\\ Integratingbothsides,wehave\\ \Rightarrow {\displaystyle \int e^y.dy}={\displaystyle \int \left(e^x+x^2\right)}dx\\ \Rightarrow e^y=e^x+\frac{x^3}{3}+c\\ \Rightarrow e^y-e^x=\frac{x^3}{3}+c\\ Hence,thecorrectoptionis\left(b\right).\end{array}$