Differential Equations

Let $l={\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}.............\left(i\right)}$

Now ${\displaystyle \int \frac{dx}{x^2+x+1}={\displaystyle \int \left(\frac{1}{x^2+x+1}\right)dx}}$

by integration by parts

=> ${\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}=\frac{4}{3\sqrt[]{2}}ta{n}^{-1}\frac{2x+1}{\sqrt[]{3}}+\frac{1}{3}\frac{\left(2x+1\right)}{\left(x^2+x+1\right)}+c}$

$\Rightarrow a=\frac{4}{3\sqrt[]{3}},b=\frac{1}{3}$

Now, $9\left(\sqrt[]{3}a+b\right)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

$L.R.=\frac{\left|3*2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

$\frac{dy}{dx}+\frac{e^y-2x}{2x^2}=0$  

$e^{-y}\frac{dy}{dx}-\frac{e^{-y}}{x}=-\frac{-1}{2x^2}$              

Put $e^{-y}=t\Rightarrow -e^{-y}\frac{dy}{dx}=\frac{dt}{dx}$  

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$              

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

 I. F. $=e^{{\displaystyle \int \frac{dx}{x}}}=e^{lnx}=x$  

Soln. tx = ${\displaystyle \int \frac{x}{2x^2}dx=\frac{1}{2}lnx+c}$  

$x=1\Rightarrow e^{-y}=\frac{1}{2}\Rightarrow y=ln2$              

$\frac{dy}{dx}+\frac{se{c}^{2}x}{tanx}y=-tanx$

IF = tan x

Solution : y tan x = x – tan x + C

$\underset{x\to 0^{+}}{lim}xy=1\Rightarrow C=1$

For $x=\frac{\pi }{4},y=\frac{\pi }{4}$

(x + y) = 4xy

$\Rightarrow \frac{dy}{dx}=\frac{4xy+4y^2-1}{1-4x^2-4xy},$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(1-4x^2-4xy\right)\left(4y+4xy\text{'}+8yy\text{'}\right)-\left(4xy+4y^2-1\right)\left(-8x-4y-4xy\text{'}\right)}{{\left(1-4x^2-4xy\right)}^2}$

$y^{2}dx+\left(x^2-xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2-xy+y^2}{y^2}=0\Rightarrow \frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2-\left(\frac{x}{y}\right)+1=0$

Put x = vy $\Rightarrow v+y\frac{dv}{dy}+v^2-v+1=0$

$\Rightarrow \frac{ydv}{dy}+v^2+1=0$

$\Rightarrow \frac{\pi }{6}+\mathcal{l}n\left|y\right|=\frac{\pi }{4}$

$\mathcal{l}n\left|y\right|=\frac{\pi }{12}$

 $\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1},x>-1$

IF = $e^{{\displaystyle \int pdx}}=\frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x^2+11x+13}{x^3+6x^2+11x+6}dn={\displaystyle \int \left(\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right)dx}}}$

$=\mathcal{l}n\left({\left(x+1\right)}^2\cdot \left(x+2\right)/\left(x+3\right)\right)$

$\frac{2x^2+11x+13}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$

$2x^2+11x+13=A\left(x+2\right)\left(x+3\right)+B\left(x+1\right)\left(x+3\right)+C\left(x+1\right)\left(x+2\right)$

x = -1

⇒ 4 = 2A ⇒ A = 2

x = -2

⇒ -1 = -B Þ B = 1

x₂ – 3 ⇒ -2 = 2c

c = -1

$y\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}={\displaystyle \int \frac{x+3}{x+1}\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}dx}$

= ${\displaystyle \int \left(x+1\right)\left(x+2\right)dx}$

= $\frac{x^3}{3}+\frac{3x^2}{2}+2x+c$

$\left(0,1\right)\Rightarrow 1\cdot \frac{2}{3}=c$

x = 1 $y\cdot \left(3\right)=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}=\frac{3}{2}+3=\frac{9}{2}$

y = $\frac{3}{2}$

$\frac{dy}{dx}=\frac{x+y-2}{x-y}$

Let x – 1 = X, y – 1 = Y

then DE: $\frac{dY}{dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$

Put y = vx

then $\frac{dY}{dX}=V+X\frac{dV}{dX}$

V + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

$X\frac{dV}{dX}=\frac{1+V}{1-V}-V$

$=\frac{1+V^2}{1-V}$

$\frac{1-V}{1+V^2}dV=\frac{dX}{X}$

$\frac{V-1}{V^2+1}dV+\frac{dX}{X}=0$

$\frac{1}{2}\mathcal{l}n\left|V^2+1\right|-ta{n}^{-1}V+\mathcal{l}n\left|X\right|=c$

$\mathcal{l}n\sqrt[]{V^2+1}\cdot X-ta{n}^{-1}V=c$

$\mathcal{l}n\left(\sqrt[]{1+{\left(\frac{Y-1}{X-}\right)}^2}\cdot \left|X-1\right|\right)ta{n}^{-1}\frac{Y-1}{X-1}=c$

$\left(2,1\right)\Rightarrow \mathcal{l}n\left(\sqrt[]{1+0}\cdot 1\right)-0=c$

c = 0

$\mathcal{l}n\sqrt[]{{\left(X-1\right)}^2+{\left(Y-1\right)}^2}=ta{n}^{-1}\frac{Y-1}{X-1}$

point (k + 1, 2) $\mathcal{l}n\sqrt[]{k^2+1}=ta{n}^{-1}\frac{1}{k}$

$\Rightarrow \frac{1}{2}\mathcal{l}n\left(k^2+1\right)=ta{n}^{-1}\frac{1}{k}$

$2x^2\frac{dy}{dx}-2xy+3y^2=0$

$\frac{dy}{dx}=\frac{y}{x}-\frac{3}{2}{\left(\frac{y}{x}\right)}^2$

Put y = vx

  $\frac{dy}{dx}=v+x\frac{dv}{dx}$

$Point\left(e,\frac{e}{3}\right)$

$\Rightarrow \frac{-e}{\left(\frac{e}{3}\right)}+\frac{3}{2}\mathcal{l}ne=C$

$C=\frac{3}{2}-3=\frac{-3}{2}$

$\frac{3}{2}\mathcal{l}n\left|x\right|\frac{-x}{y}+\frac{3}{2}=0$

$x=1\Rightarrow y=\frac{2}{3}$              

$b_n={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{co{s}^{2}nx}{sinx}dx}$

$={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1+cos2nx}{2sinx}dx}$

$b_n-b_{n-1}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{cos\left(2nx\right)-cos\left(2\left(n-1\right)x\right)}{2sinx}}dx$

$={\frac{cos\left(2n-1\right)x}{\left(2n-1\right)}]}_0^{\pi /2}=\frac{1}{2n-1}\left[0-1\right]=\frac{-1}{2n-1}$

$b_{n-1}-b_n=\frac{1}{2n-1}$

$b_1-b_2=\frac{1}{3}\left(A\right)-\frac{1}{5},......$

$b_2-b_3=\frac{1}{3}\left(B\right)-5,-7,-9,.....$

$b_3-b_4=\frac{1}{7}\left(C\right)$