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.

15. Option (ii) Low temperature is correct since at sufficiently low temperature, the thermal energy is low, so the intermolecular forces bring the molecules of a substance closer so that they cling to one another and occupy fixed positions. They keep on vibrating about their fixed positions. Such conditions favours the existence of the substance in solid state. 

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

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

14. On heating the amorphous substance it gets changed to the crystalline form at some temperature. This is due to the process called crystallization. As on heating at some temperature it may become crystalline since slow heating and cooling over a longer period of time makes these changes.

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.

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