Class 12th

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.

Given $f{x}^n$ is $y=e^x+1$

Differentiating with $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of $y^{\left|\right|}$ and $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

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

86. To prove $\frac{d}{dx}(uv·w)=\frac{du}{dx}v·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}.$

By repeating application of produced rule

$=\frac{d}{dx}(uv:u)$

$=u\frac{d}{dx}(u·w)+v·w\frac{dy}{dx}$

$u\left\{v\frac{dw}{dx}+w\frac{dv}{dx}\right\}+\frac{dy}{dx}v:w.$

$=u·v·\frac{dw}{dx}+u·\frac{dv}{dx}w+\frac{dy}{dx}·v·w$

$=\frac{dy}{dx}u·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}$ = R*H*S*

By togarith differentiating,

Let y = u v w

Taking log, log y = log u + log v + log w

Differentiating w r t 'x'

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{4}\frac{dy}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]$

$\Rightarrow \frac{d}{dx}(uvw)=uvw·\left[\frac{1}{u}\frac{dy}{dx}+\frac{1}{v}\frac{du}{dx}+\frac{1}{w}\frac{dur}{dx}\right]$

$=\frac{dy}{dx}v·w+u\frac{dv}{dx}·w+uv·\frac{dw}{dx}$