Ncert Solutions Maths class 12th

The given D.E. is

$\frac{dy}{dx}+\frac{y}{x}=x^2$ Which is in the form $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{1}{x}=\&Q=x^2$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{2}dx}}=e^{logx}=x\left\{?e^{logx}=x\right\}$

Thus, the general solution is

$\begin{array}{l}y\times I.F={\displaystyle \int Q\times I.Fdx+c}\\ y.x={\displaystyle \int x^2.xdx+c}\\ =xy={\displaystyle \int x^{3}dx+c}\\ =xy=\frac{x^4}{4}+c\end{array}$

135. Given, $f(x)=|x{|}^3=\{\begin{array}{cc}\hfill x^3\hfill & \hfill if x\ge 0\hfill \\ \hfill -x^3\hfill & \hfill if x<0\hfill \end{array}$

For $x\ge 0,f(x)=|x{|}^3=x^3$

and $\begin{array}{cc}\hfill f^{\prime }(x)=3x^2\hfill & \hfill f^{\prime \prime }(x)=6x\hfill \end{array}$

For $x<0,f(x)=|x{|}^3={(-x)}^3=-x^3.$

so, $\begin{array}{cc}\hfill f^{\prime }(x)=-3x^2\hfill & \hfill f^{\prime \prime }(x)=-6x\hfill \end{array}$

Hence, $f^{\prime \prime }(x)=\{\begin{array}{cc}\hfill 6x,\hfill & \hfill if x\ge 0\hfill \\ \hfill -6x,\hfill & \hfill if x<0\hfill \end{array}$

Given, D.E. is

$\frac{dy}{dx}+3y=e^{-2x}$ which is of the form

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

Where $\begin{array}{l}P=3\\ \&Q=e^{-2x}\end{array}$

So, I.F $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 3dx}}=e^{3x}$

So, the solution is $=y\times I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

$\begin{array}{l}=y\times e^{3x}={\displaystyle \int e^{-2x}.}{e}^{3x}dx+c\\ =e^{3x}y={\displaystyle \int e^{x}dx+c}\\ =e^{3x}y=e^x+c\\ =y=\frac{e^x}{e^{3x}}+\frac{c}{e^{3x}}\\ =y=e^{-2x}+c{e}^{-3x}\end{array}$

Is the required general solution.

The given D.E. is

$\frac{dy}{dx}+2y=2sinx$ which is of form $\frac{dy}{dx}+Px=Q$

We have, P = 2

$Q=sinx$

So, I.F. $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 2dx}}=e^{2x}$

The solution is $y\times I.F={\displaystyle \int Q\times \left(I.F\right)dx+c}$

$\begin{array}{l}y{e}^{2x}={\displaystyle \int sinx{e}^{2x}dx+c}\\ =y{e}^{2x}=I+c---------(1)\\ Where,I={\displaystyle \int sinx{e}^{2x}}\\ =sinx{\displaystyle \int e^{2x}-{\displaystyle \int \left(\frac{d}{dx}sinx{\displaystyle \int e^{2x}dx}\right)dx}}\\ =sinx\frac{e^{2x}}{2}-\frac{1}{2}\left[{\displaystyle \int cosx{e}^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cos{\displaystyle \int e^{2x}dx-{\displaystyle \int \left(\frac{d}{dx}cosx{\displaystyle \int e^{2x}dx}\right)dx}}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cosx\frac{e^{2x}}{2}\frac{1}{2}{\displaystyle \int \left(-cosx\right)e^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}{\displaystyle \int sinx{e}^{2x}dx}\end{array}$

$\begin{array}{l}=I=e^{2x}\frac{sinx}{2}-e^{2x}\frac{cosx}{4}-\frac{1}{4}1\\ =I+\frac{1}{4}I=2e^{2x}\frac{sinx-e^{2x}cosx}{4}\\ =\frac{5}{4}I=e^{2x}\frac{\left(2sinx-cosx\right)}{4}\\ =I=e^{2x}\frac{\left(2sinx-cosx\right)}{5}\end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l}y{e}^{2x}=\frac{e^{2x}}{5}\left(2sinx-cosx\right)+c\\ =y=\frac{\left(2sinx-cosx\right)}{5}+c{e}^{2x}\end{array}$

Is the required solution.

134. Given, $x=a(cost+tsint)$ and $y=a(sint-tcost).$

Differentiating w r t. ‘t’ we get,

$\frac{dx}{dt}=a\frac{d}{dt}(cost+tsint).$

$=a\left(-sint+t\frac{d}{dt}sint+sint\frac{dt}{dt}\right).$

$=a(-sint+tcost+sint)=atcost$

$\frac{dy}{dt}=\frac{ad}{dt}(csint-tcost)$

$=a\left(cost-t\frac{d}{dt}cost-cot\frac{dt}{dt}\right)$

$=a(cost+tsint-cost)=atsint$

$\frac{dy}{dx}=\frac{dy/dt}{dx/at}=\frac{atsint}{atcost}=tant$

So, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(tant)=\frac{d}{dt}(tant)\cdot \frac{dt}{dx}.$

$=se{c}^{2}t\times \frac{dt}{dx}.$

$=se{c}^{2}t\times \frac{1}{(dx/dt)}$

$=se{c}^{2}t\times \frac{1}{ at cost}$

$=\frac{se{c}^{3}t}{at}$

133.  Given, $cosy=xcos(a+y).$

$x=\frac{cosy}{cos(a+y)}$

Differentiating w r t ‘y’ we get,

$\frac{dx}{dy}=\frac{d}{dy}\left(\frac{cosy}{cos(a+y)}\right).$

$=\frac{cos(a+y)\frac{d}{dy}cosy-cosy\frac{d}{dy}cos(a+y)}{co{s}^2(a+y).}$

$=\frac{cos(a+y)(-siny)-cosy(-sin(a+y))}{co{s}^2(a+y).}$

$=\frac{-cos(a+y)siny+sin(a+y)cosy}{co{s}^2(a+y)}$

$=\frac{sin(a+y)cosy-cos(a+y)siny.}{co{s}^2(a+y)}$

$\frac{dx}{dy}=\frac{sin(a+y-y)}{co{s}^2(a+y)}\left\{\begin{array}{rr}\hfill ?sin(A-B)=sinAcosB& \hfill -cosAsinB\end{array}\right\}$

So, $\frac{dy}{dx}=\frac{co{s}^2(a+y)}{sina}$

132. Given, ${(x-a)}^2+{(y-b)}^2=c^2.$

Differentiating w r t ‘x’ we get

$\frac{d}{dx}{(x-a)}^2+\frac{d}{dx}{(y-b)}^2=\frac{d}{dx}{c}^2$

$\Rightarrow 2(x-a)+2(y-b)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{2(x-a)}{2(y-b)}=-\frac{(x-a)}{(y-b)}$

Again, $\frac{d^{2}y}{d{x}^2}=-\left\{\frac{(y-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(y-b)}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)-(x-a)\frac{dy}{dx}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)+(x-a)\frac{(x-a)}{(y-b)}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{{(y-b)}^2+{(x-a)}^2}{{(y-b)}^3}\right\}$

$=-\frac{c^2}{{(y-b)}^3}\left\{?{(x-a)}^2+{(y-b)}^2=c^2\right\}$

Then, L.H.S = $\frac{{\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\}}^{3/2}}{\frac{d^{2}y}{d{x}^2}}=\frac{{\left\{1+\frac{{(x-a)}^2}{{(y-b)}^2}\right\}}^{3/2}}{\frac{-c^2}{{(y-b)}^2}}$

$=\frac{{\left\{{(y-b)}^2+{(x-a)}^2\right\}}^{3/2}}{{(y-b)}^3}\times \frac{{(y-b)}^3}{-c^2}$

$=\frac{c^{2\times 3/2}}{-c^2}=\frac{c^3}{-c^2}=-c$ Where c is a constant and is independent of a and b.

131. Kindly go through the solution

130. Kindly go through the solution

129. Given, $y=12(1-cost). x=10(t-sint).$

Differentiating w r t ‘t’ we get,

$\frac{dy}{dt}=12\frac{d}{dt}(1-cost)=12(0-(-sint))=12sint.$

$\frac{dx}{dt}=10\frac{d}{dt}(t-sint)=10(1-cost).$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

$=\frac{12sint}{10(1-cost)}=\frac{12sint}{10[1-cost]}$

$=\frac{12\times 2sint/tcost/2}{10\times 2si{n}^{2}t/2}$

$\left\{\begin{array}{cc}\hfill ?sin2\theta =2sin\theta cos\theta \hfill & \hfill cos2\theta =1-2si{n}^2\theta \hfill \end{array}\right\}$ $=\frac{6}{5}\frac{cost/2}{sint/2}=\frac{6}{5}cott/2$