Ncert Solutions Maths class 12th

The given D.E. is

$\left(1-y^2\right)\frac{dx}{dy}+yx=\propto y\left(-1<y<1\right)$

$\frac{dx}{dy}+\frac{y}{1-y^2}*x=\frac{\propto y}{1-y^2}$ which is of form

$\begin{array}{l}\frac{dx}{dy}+Px=Q\&P=\frac{y}{1-y^2}\&Q=\frac{\propto y}{1-y^2}\\ {\displaystyle \int pdx={\displaystyle \int \frac{y}{1-y^2}dx=-\frac{1}{2}{\displaystyle \int \frac{-2y}{1-y^{2}dx}}}}\end{array}$

$\begin{array}{l}=-\frac{1}{2}log\left|1-y^2\right|\\ =log{\left[1-y^2\right]}^{-\frac{1}{2}}\end{array}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log{\left[1-y^2\right]}^{-\frac{1}{2}}}\\ ={\left[1-y^2\right]}^{-\frac{1}{2}}$

$\therefore$ option (D ) is correct.

The given D.E is

$\begin{array}{l}x\frac{dy}{dx}-y=2x^2\\ \frac{dy}{dx}-\frac{1}{x}y=2x\end{array}$

Which is of form $\frac{dy}{dx}+Py=Q$

So,  $P=-\frac{1}{x}$

$\therefore I.E=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{x}dx}}=e^{log-x}=e^{log-x^{-1}}=x^{-1}=\frac{1}{x}$

$\therefore$ Option (c) is correct

We know that slope of tangent to the curve is $\frac{dy}{dx}$

$\therefore x+y=\frac{dy}{dx}+5$

$\frac{dy}{dx}-y=x-5$ Which has form $\frac{dy}{dx}+Px=Q$

$where,P=1\&Q=x-5$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -1dx}}=e^{-x}$

Thus the solution has the form

$\begin{array}{l}y{e}^{-x}={\displaystyle \int \left(x-5\right)e^{-x}dx+c}\\ ={\displaystyle \int x{e}^{-x}dx-{\displaystyle \int 5e^{-x}dx+c}}\\ =y{e}^{-x}=I+5e^{-x}+c\\ where,I={\displaystyle \int x{e}^{-x}dx}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int e^{-x}dxdx}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx}\\ =-x{e}^{-x}-e^{-x}\end{array}$

$\begin{array}{l}\therefore y{e}^{-x}=-x{e}^{-x}-e^{-x}+5e^{-x}+c\\ =y{e}^{-x}=-x{e}^{-x}+4e^{-x}+c\\ =y=-x+4+\frac{c}{e^{-x}}\\ =y+x=4+c{e}^x\end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l}2+0=4c{e}^0\\ 2-4=c\\ c=-2\end{array}$

$\therefore$ The particular solution is

$y+x=4-2e^x$

We know the slope of tangent to curve is $\frac{dy}{dx}$ .

$\therefore$ $\frac{dy}{dx}=x+y$

$=\frac{dy}{dx}-y=x$ which has form $\frac{dy}{dx}+Py=Q$

So, $P=-1\&Q=x$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -dx}}=e^{-x}$

Thus the solution is of the form .

$\begin{array}{l}y*e^{-x}={\displaystyle \int x.e^{-x}dx+c}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int e^{-x}dxdx+c}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx+c}\\ =y{e}^{-x}=-x{e}^{-x}-e^{-x}+c\\ =y=-x-1+\frac{c}{e^{-x}}\\ =y+x+1=c{e}^x\end{array}$

Given, the curve passes through origin (0,0) i.e, $y=0,when,x=0$

$0+0+1=c{e}^0=c=1$

$\therefore$ Thus, equation of the curve is

$y+x+1=e^x$

The given D.E. is

$\frac{dy}{dx}-3ycotx=sin2x$

$\frac{dy}{dx}-3cotx.y=sin2x$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=-3cotx\&Q=sin2x$

$\therefore I.F=e^{-{\displaystyle \int Pdx}}=e^{{\displaystyle \int -3cotxdx}}=e^{-3{\displaystyle \int cotxdx}}=e^{-3log\left|cotxdx\right|}=e^{log{\left(sin\right)}^{-3}}=\frac{1}{si{n}^{3}x}$

Thus the solution is of the form.

$\begin{array}{l}y*\frac{1}{si{n}^{3}x}={\displaystyle \int sin2x.\frac{1}{si{n}^{3}x}dx+c}\\ ={\displaystyle \int \frac{2sinxcosx}{si{n}^{3}x}}dx+c\left\{?sin2x=2sinxcosx\right\}\\ ={\displaystyle \int 2cosecxcotxdx+c}\\ =-2cosecx+c\\ =\frac{y}{si{n}^{3}x}=\frac{-2}{sinx}+c\\ =-2y=-2si{n}^{2}x+csi{n}^{3}x\end{array}$

Given, $y=2,when,x=\frac{\pi }{2}$

$\begin{array}{l}2=-2si{n}^2\frac{1}{2}+csi{n}^3\frac{\pi }{2}\\ =2=-2+c\\ =e=2+2=4\end{array}$

$\therefore$ The particular solution is, $y=-2si{n}^{2}x+4si{n}^{3}x$

The given D.E. is

$\left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$

Which is of form 

$\frac{dy}{dx}+Px=Q$

So, $P=\frac{2x}{1+x^2}\&Q=\frac{1}{{\left(1+x^2\right)}^2}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x}{1+x^2}dx=log\left|1+x^2\right|}}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is if form,

$y*\left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

$\begin{array}{l}y\left(1+x^2\right)={\displaystyle \int \frac{1}{{\left(1+x^2\right)}^2}*\left(1+x^2\right)dx+c}\\ ={\displaystyle \int \frac{1}{\left(1+x^2\right)}dx+c}\\ y*\left(1+x^2\right)=ta{n}^{-1}x+c\end{array}$

Given, $y=0,when,x=1$

$\begin{array}{l}0=ta{n}^{-1}1+c\\ c=ta{n}^{-1}1=-\frac{\pi }{4}\end{array}$

$\therefore$ The particular solution is

$y\left(1+x^2\right)=ta{n}^{-1}x-\frac{\pi }{4}$

The given D.E. is

$\frac{dy}{dx}+2ytanx=sinx$

$\Rightarrow \frac{dy}{dx}+\left(2tanx\right)y=sinx$ Which is of form $\frac{dy}{dx}+Px=Q$

So, $P=2tanx\&Q=sinx$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int 2tanxdx}}=e^{2log\left|secx\right|}=e^{logse{c}^2}=se{c}^2$

Thus the solution is of the form $y*\left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

$\begin{array}{l}\Rightarrow y.se{c}^{2}x={\displaystyle \int sinx.se{c}^{2}xdx+c}\\ =\frac{sinx}{co{s}^2}dx+c\\ ={\displaystyle \int tanx.secxdx+c}\\ =yse{c}^2=secx+c\\ =y=\frac{1}{secx}+\frac{c}{se{c}^{2}x}=cosx+cco{s}^2\\ =y=cosx+cco{s}^{2}x\end{array}$

Given, $y=0,Whenx=\frac{\pi }{3}$

$\begin{array}{l}=0=cos\frac{\pi }{3}+cco{s}^2\frac{\pi }{3}\left\{\frac{c}{4}=-\frac{1}{2},c=-\frac{4}{2},c=-2\right\}\\ =0=\frac{1}{2}+c{\left(\frac{1}{2}\right)}^2\\ =0=\frac{1}{2}+\frac{c}{4}\end{array}$

C = -2

$\therefore$ The particular solution is

$y=cosx-2co{s}^{2}x$

The given D.E. is

$\begin{array}{l}\left(x+3y^2\right)\frac{dy}{dx}=y\\ \Rightarrow \left(x+3y^2\right)dy=ydx\\ \Rightarrow y\frac{dx}{dy}=x+3y^2\\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}+3y\end{array}$

$\Rightarrow \frac{dx}{dy}-\frac{1}{y}.x=3y$ Which is form $\frac{dx}{dy}+Px=Q$

So, $P=-\frac{1}{y}\&Q=3y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -\frac{1}{y}dy}}=e^{-log\left|y\right|}=e^{log{y}^{-1}}=y^{-1}=\frac{1}{y}$

Thus the solution is of the form.

$\begin{array}{l}x*\frac{1}{y}={\displaystyle \int 3y.\frac{1}{y}dy+c}\\ =\frac{x}{y}={\displaystyle \int 3dy+c}\\ =\frac{x}{y}=3y+c\\ =x=3y^2+cy\end{array}$

The given D.E. is

$\begin{array}{l}ydx+\left(x-y^2\right)dy=0\\ =y\frac{dx}{dy}+x-y^2=0\\ =\frac{dx}{dy}+\frac{x}{y}-y=0\end{array}$