Class 12th

The given D.E. is $\frac{xdy}{dx}-y+xsin\left(\frac{y}{x}\right)=0$

$\begin{array}{l}\Rightarrow \frac{xdy}{dx}=y-xsin\left(\frac{y}{x}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-xsin\left(\frac{y}{x}\right)}{x}=\frac{y}{x}-sin\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx=\frac{y}{x}=v$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E.

Then, $v+x\frac{dv}{dx}=v-sinv$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=-sinv\\ \Rightarrow \frac{dv}{sinv}=-\frac{dx}{x}\\ \Rightarrow cosecvdv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int cosecvdv=}{\displaystyle \int -\frac{dx}{x}}\\ \Rightarrow log\left|cosecv-cotv\right|=-logx+logc\\ \Rightarrow log\left|cosecv-cotv\right|=log\frac{c}{x}\\ \Rightarrow cosecv-cotv=\frac{c}{x}\end{array}$

Putting back $v=\frac{y}{x}$ we get,

$\begin{array}{l}cosec\frac{y}{x}-cot\frac{y}{x}=\frac{c}{x}\\ \Rightarrow \frac{1}{sin\frac{y}{x}}-\frac{cos\frac{y}{x}}{sin\frac{y}{x}}=\frac{c}{x}\end{array}$

$\Rightarrow x\left[1-cos\frac{y}{x}\right]=csin\left(\frac{y}{x}\right)$ is the solution of the D.E.

The given D.E is

. $\begin{array}{l}\left\{xcos\left(\frac{y}{x}\right)+ysin\left(\frac{y}{x}\right)\right\}ydx=\left\{ysin\left(\frac{y}{x}\right)-xcos\left(\frac{y}{x}\right)\right\}xdy\\ \Rightarrow \frac{dy}{dx}=\frac{\left\{xcos\frac{y}{x}+ysin\frac{y}{x}\right\}y}{\left\{ysin\frac{y}{x}-xcos\frac{y}{x}\right\}x}=\frac{xycos\frac{y}{x}+y^{2}sin\frac{y}{x}}{xysin\frac{y}{x}-x^{2}cos\frac{y}{x}}\end{array}$

$\Rightarrow \frac{\frac{y}{x}cos\frac{y}{x}+{\left(\frac{y}{x}\right)}^2-sin\frac{y}{x}}{\left(\frac{y}{x}\right)sin\frac{y}{x}-cos\frac{y}{x}}$ {Dividing numerator and denominator by $x^2$ }

$=f\left(\frac{y}{x}\right)$

Hence, the given D.E is homogenous.

Let, $y=vx=\frac{y}{x}=v$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E.

Then, $v+x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv}{vsinv-cosv}$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=\frac{vcosv+v^{2}sinv}{vsinv-cosv}-v=\frac{vcosv+v^{2}sinv-v^{2}sinv+vcosv}{vsinv-cosv}\\ \Rightarrow \frac{vcosv}{vsinv-cosv}\\ \Rightarrow \left(\frac{vsinv-cosv}{vcosv}\right)dv=\frac{2dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{vsinv-cosv}{vcosv}dv=2{\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int tanvdv-{\displaystyle \int \frac{1}{v}dv=2log\left|x\right|+}}log\left|c\right|\\ \Rightarrow log\left|secv\right|-log\left|v\right|=log{x}^2+logc\\ \Rightarrow log\left|\frac{secv}{v}\right|=logc{x}^2\\ \Rightarrow \frac{secv}{v}=c{x}^2\\ \Rightarrow secv=c{x}^{2}v\end{array}$

Putting back $\Rightarrow v=\frac{y}{x}=c{x}^2\frac{y}{x}=cxy$

$\begin{array}{l}sec\frac{y}{x}=c{x}^2\frac{y}{x}=cxy\\ \Rightarrow \frac{1}{cos\frac{y}{x}}=cxy\\ \Rightarrow \frac{1}{c}=xycos\frac{y}{x}\end{array}$

$\Rightarrow xycos\frac{y}{x}=c_1$ where $c_1=\frac{1}{c}$ 

115.

Solution :
The given function is $f\left(x\right)=x^2-4x-3$ f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

$\begin{array}{l}f\left(1\right)=1^2-4*1-3,f\left(4\right)=4^2-4*4-3\\ \therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{-3-\left(-6\right)}{3}=\frac{3}{3}=1\end{array}$

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f' (c) = 1

$\begin{array}{l}f\text{'}\left(c\right)=1\\ \Rightarrow 2c-4=1\\ \Rightarrow c=\frac{5}{2},where,c=\frac{5}{2}\in \left(1,4\right)\end{array}$

Hence, Mean Value Theorem is verified for the given function.

Kindly go through the solution

Integrating both sides,

114. Solution :
It is given that f: [-5,5]? R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [?5, 5].

(b) f is differentiable on (?5, 5).

Therefore, by the Mean Value Theorem, there exists c? (?5, 5) such that

$\begin{array}{l}{f}^{\text{'}}\left(c\right)\frac{f\left(5\right)?f\left(?5\right)}{5?\left(?5\right)}\\ ?10f^{\text{'}}\left(c\right)=f\left(5\right)?f\left(?5\right)\end{array}$

It is also given that f' (x) does not vanish anywhere.

$\begin{array}{l}?f^{\text{'}}\left(c\right)?0\\ ?10f^{\text{'}}\left(c\right)?0\\ ?f\left(5\right)?f\left(?5\right)?0\\ ?f\left(5\right)?f\left(?5\right)\end{array}$

Hence, proved.

The given D.E. is

$\begin{array}{l}{x}^2\frac{dy}{dx}=x^2-2y^2++xy\\ \Rightarrow \frac{dy}{dx}=\frac{x^2-2y^2++xy}{x^2}=1-\frac{2y^2}{x^2}+\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=1-2{\left(\frac{y}{x}\right)}^2+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the D.E. is homogenous $f{x}^n$

Let, $y=vx.=\frac{y}{x}=v$ so that, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Thus, $v+x\frac{dv}{dx}=1-2v^2+v$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=1-2v^2\\ \Rightarrow \frac{dv}{1-2v^2}=\frac{dx}{x}\end{array}$

Integrating both sides,

The Given D.E. is

$\begin{array}{l}\left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow 2*ydy=-\left(x^2-y^2\right)dx\\ =\frac{dy}{dx}=-\frac{\left(x^2-y^2\right)}{2xy}=\frac{\left(y^2-x^2\right)}{2xy}\\ =\frac{1}{2}\frac{\left(\frac{y^2-x^2}{x^2}\right)}{\left(\frac{xy}{x^2}\right)}\\ =\frac{1}{2}\frac{\left[{\left(\frac{y}{x}\right)}^2-1\right]}{\left(\frac{y}{x}\right)}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx\Rightarrow \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

$\begin{array}{l}\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}\left[\frac{v^2-1}{v}\right]\\ \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=\frac{-1-v^2}{2v}\\ \Rightarrow \left(\frac{2v}{1+v^2}\right)dv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

Putting back $v=\frac{y}{x}$ we get,

$\begin{array}{l}\Rightarrow x\left[\frac{y^2}{x^2}+1\right]=c\\ \Rightarrow \frac{y^2+x^2}{x}=c\end{array}$

$\Rightarrow y^2+x^2=xc$ is the required solution.

The Given D.E. is $\left(x-y\right)dy-\left(x+y\right)dx=0$

$\begin{array}{l}\Rightarrow \left(x-y\right)dy=\left(x+y\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}=\frac{x\left(1+\frac{y}{x}\right)}{x\left(1-\frac{y}{x}\right)}=\frac{1+\frac{y}{x}}{1-\frac{y}{x}}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx=\frac{y}{x}=v,so,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then, $\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-1+v^2}{1-v}=\frac{1+v^2}{1-v}\\ \Rightarrow \left[\frac{1-v}{1+v^2}\right]dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1-v}{1+v^2}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1}{1+v^2}dv-\frac{1}{2}{\displaystyle \int \frac{2v}{1+v^2}dv=logx+c}}\\ \Rightarrow ta{n}^{-1}v-\frac{1}{2}log\left(1+v^2\right)=logx+c\end{array}$

Putting back $v=\frac{y}{x},weget,$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|1+\frac{y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|\frac{x^2+y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}\left[log\left(x^2+y^2\right)-log{x}^2\right]=logx+c\end{array}$

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