Differential Equations Applications

I = ∫ (e? (x²+1)/ (x+1)² dx = f (x)e? + c
I = ∫ (e? (x²-1+1+1)/ (x+1)² dx
I = ∫e? [ (x-1)/ (x+1) + 2/ (x+1)² ] dx
for x = 1
f' (1) = 12/24 - 12/16 = 3/4

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

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

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

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

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

Put y = vx

  $\frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\frac{dy}{dx}-y=x,$  linear differential equation.

IF = e^-x

$y.e^{-x}={\displaystyle \int x{e}^{-x}dx=-e^{-x}\left(x+1\right)+C}$                

$y=-\left(x+1\right)+C.e^x$                

$y_2\left(x\right)=-\left(x+1\right)+2e^x$                

y₂ > y₁, no solution.

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

x, y > 0, y(1) = 1

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

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

$=\frac{lo{g}_e\left(2^y-1\right)}{lo{g}_{e}2}=-\frac{lo{g}_e\left(2^x-1\right)}{lo{g}_{e}2}+\frac{lo{g}_{e}c}{lo{g}_{e}2}$  

Taking log of base 2.

$\therefore$  y = 2 – log₂ 

  $l={\displaystyle \int \frac{e^x\left(x^2+1\right)}{{\left(x+1\right)}^2}dx=f\left(x\right)e^X+c}$

$l=\text{?}{\displaystyle \int \frac{e^x\left(x^2-1+1+1\right)}{{\left(x+1\right)}^2}dx}$

 $={\displaystyle \int e^x}\left[\frac{x-1}{x+1}+\frac{2}{{\left(x+1\right)}^2}\right]dx$

 for x = 1

$f\text{'}\text{'}\text{'}\left(1\right)=\frac{12}{24}=\frac{12}{16}=\frac{3}{4}$                

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

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 Þ A = 2

x = -2

-> -1 = -B Þ B = 1

x₂ – 3 Þ -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 + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

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

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

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

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

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

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

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

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

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