Ncert Solutions Maths class 12th

The given D.E.is

$\begin{array}{l}co{s}^{2}x\frac{dy}{dx}+y=tanx\\ =\frac{dy}{dx}+\frac{1}{co{s}^{2}x}y=\frac{tanx}{co{s}^{2}x}\end{array}$

$=\frac{dy}{dx}+se{c}^{2}xy=se{c}^{2}xtanx$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=se{c}^{2}x\&Q=se{c}^{2}xtanx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int se{c}^{2}dx}}=e^{tanx}$

Thus, the general solution is of the form.

$y.e^{tanx}={\displaystyle \int se{c}^{2}xtanx.}{e}^{tanx}dx+c$

Let, $tanx=t=se{c}^{2}xdx=dt$

$\begin{array}{l}=y{e}^t={\displaystyle \int t.}{e}^{t}dt+c\\ =t{\displaystyle \int e^{t}dt-{\displaystyle \int \frac{d}{dt}t{\displaystyle \int e^{t}dt.dt+c}}}\\ =t{e}^t-{\displaystyle \int e^{t}dt+c}\\ =t{e}^t-e^t+c\\ =e^t\left(t-1\right)+c\end{array}$

$\begin{array}{l}\Rightarrow y{e}^{tanx}=e^{tanx}\left(tanx-1\right)+c\\ y=\left(tanx-1\right)+c{e}^{-tanx}\end{array}$

136. Given, $sin(A+B)=sinAcosB+cosAsinB$

Differentiating w r t. 'x' we get,

$\frac{d}{dx}sin(A+B)=\frac{d}{dx}(sinAcosB+cosAsinB)$

$\to cos(A+B)-\frac{d}{dx}(A+B)=sinA\frac{d}{dx}cosB+cosB\frac{d}{dx}sinA+cosA\frac{d}{dx}sinB+sinB\frac{d}{dx}cosA$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=-sinAsinB\frac{dB}{dx}+cosBcosA\frac{dA}{dx}+cosAcosB\frac{dB}{dx}-sinAsinB\frac{dA}{dx}$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dt}\right)=cosAcosB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)-sinAsinB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)$

$=(cosAcosB-sinAsinB)\left(\frac{dA}{dx}+\frac{dB}{dx}\right).$

$\to cos(A+B)=cosAcosB-sinAsinB.$

The given D.E.is

$\frac{dy}{dx}+\left(secx\right)y=tanx$ Which is in the form $\frac{dy}{dx}+Py=Q$

So, $P=secx\&Q=tanx$

$\therefore$ $\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int secxdx}}=e^{log\left|secx+tanx\right|}=secx+tanx$

Thus, the general solution is ,

$\begin{array}{l}y*I.F={\displaystyle \int Q*I.Fdx+c}\\ =y*\left(secx+tanx\right)={\displaystyle \int tanx}\left(secx+tanx\right)dx+c\end{array}$

$\begin{array}{l}={\displaystyle \int \left(tanxsecx+ta{n}^{2}x\right)dx+c}\\ ={\displaystyle \int \left(tanxsecx+se{c}^2-1\right)}dx+c\\ =sec+tanx-x+c\end{array}$

$=\left(secx+tanx\right)y=secx+tanx-x+c\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

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

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

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 $=y*I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

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

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 $y*I.F={\displaystyle \int Q*\left(I.F\right)dx+c}$

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

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

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

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

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

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

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

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

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