Ncert Solutions Maths class 12th

Given $y-cosy=x$

Differentiate w.r.t 'x' we get

$\begin{array}{l}\frac{dy}{dx}-\left(-siny\right)\frac{dy}{dx}=\frac{dx}{dx}\\ \Rightarrow \frac{dy}{dx}\left[1+siny\right]=1\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+siny}=y^{|}\end{array}$

So, L.H.S of given D.E $=\left(ysiny+cosy+x\right)y^{|}$

$\begin{array}{l}=\left(ysiny+cosy+y-cosx\right)\left[\frac{1}{1+siny}\right]\\ =\frac{y\left(1+siny\right)}{\left(1+siny\right)}=y=R.H.S\end{array}$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

89. Given, x = sin t and y = cos2t. differentiation w r t. 't' we get,

$\begin{array}{l}\frac{dx}{dt}=cost\frac{dy}{dt}=\left(-sin2t\right)\frac{d2t}{dt}\\ -t\\ -2\left(2sintcost\right)\\ -tt\end{array}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}$

$=\frac{-4sintcost}{cost}$

= -4 sin t

Given, $xy=logy+c$

Differentiate w.r.t. x we have

$\begin{array}{l}x\frac{dy}{dx}+y\frac{dx}{dx}=\frac{d}{dx}logy+\frac{d}{dx}C\\ \Rightarrow x\frac{dy}{dx}+y=\frac{1}{y}\frac{dy}{dx}+0\\ \Rightarrow x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}\left[x-\frac{1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}\left[\frac{xy-1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{xy-1}=\frac{\left(-1\right)*-y^2}{\left(-1\right)*\left(xy-1\right)}=\frac{y^2}{1-xy}\\ \therefore y^{|}=\frac{y^2}{1-xy}\end{array}$

Hence, y is a Solution of the given D.E

Given,  $y=xsinx$

So,  $y^{|}=x\frac{d}{dx}sinx+sinx\frac{dx}{dx}=xcosx+sinx$

Now, L.H.S of the given D.E $=x{y}^{|}$

$=x\left(xcosx+sinx\right)$

$=x^{2}cosx+xsinx$

88. Kindly go through the solution

Given,  $y=Ax:$

So,  $y^{|}=\frac{Adx}{dx}=A$

Putting value of $y^{|}$ in L.H.S. of the given D.E.

L.H.S= $x{y}^{|}=xA=Ax=y$ =R.H.S

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given, y= √1 + x²

87. Given, x = 2at2 and y = at4. Differentiation w r t we get,

$\frac{dx}{dt}=4at.$ and $\frac{dy}{dt}=4a{t}^3.$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{4a{t}^3}{4at}=t^2.$

Given,  $f{x}^n$ is $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given,  $f{x}^n$ is $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.