Maths

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

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

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

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

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

15. ${\left[3x^2-2ax+3a^2\right]}^3$

= ${\left[3x^2+a\left(-2x+3a\right)\right]}^3$

We know that by binomial theorem,

${(a+b)}^3$ = $a^3+b^3+3ab\left(a+b\right)$

= $a^3+b^3+3a^{2}b+3a{b}^2$

Then,

${\left[3x^2+a\left(-2x+3a\right)\right]}^3$

= (3x²)³ + ${\left[a\left(-2x+3a\right)\right]}^3$ + $\left[3(3x^2){}^2 a\left(-2x+3a\right)\right]$ + $\left[ 3(3x^2){\left\{a\left(-2x+3a\right)\right\}}^2\right]$

= 27x⁶ + $\left[a^3{\left(-2x+3a\right)}^3\right]$ + $\left[3(9x^4)\left(-2ax+3a^2\right)\right]$ + $\left[3(3x^2)\left\{a^2{\left(3a-2x\right)}^2\right\}\right]$

= 27x⁶ + $\left[a^3\left\{-8x^3+27a^3+3\left(4x^2\right)\left(3a\right)+3\left(-2x\right)\left(9a^2\right)\right\}\right]$ + $\left[-54a{x}^5+81a^2{x}^4\right]$ + [ $\left(9a^2{x}^2\right)$ $\left(9a^2+4x^2-12ax\right)$ ]

= 27x⁶ + [ $-8a^3{x}^3+27a^6+36a^4{x}^2-54a^{5}x$ ] + [ $-54a{x}^5+81a^2{x}^4$ ] + [ $(81a^4{x}^2+36a^2{x}^4-108a^3{x}^3$ ]

= 27x⁶ $- 8a^3{x}^3+27a^6+36a^4{x}^2-54a^{5}x$ $- 54a{x}^5+81a^2{x}^4$ + $81a^4{x}^2+36a^2{x}^4-108a^3{x}^3$

= 27x⁶– 54ax⁵ $+ 117a^2{x}^4$ $- 116a^3{x}^3$ + $117a^4{x}^2$ $- 54a^{5}x$ $+ 27a^6$