Differential Equations

$\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$                

$y\frac{dy}{dx}=\frac{y^2-x}{5x+y^2}$            

Let y² = t

$\frac{1}{2}.\frac{dt}{dx}=\frac{t-x}{5x=t}$  

Now substitute, t = yx

$\frac{dt}{dx}=v+x\frac{dv}{dx}$

  $\left|\frac{{\left(v+1\right)}^4}{{\left(v+2\right)}^3}\right|=\frac{C}{x}$

$\left|{\left(y^2+x\right)}^4\right|=C\left|{\left(y^2+2x\right)}^3\right|$    

$\frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}=e^{2x}-\frac{6e^{-x}}{2+9e^{-2x}}$

${\displaystyle \int dy={\displaystyle \int e^{2x}dx-3{\displaystyle \int \frac{e^{-z}}{\underset{put{e}^{-x}=t}{\underset{?}{1+\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)}}}}dx}}$

= $y=\frac{e^{2x}}{2}=\sqrt[]{2}ta{n}^{-1}\left(\frac{3e^{-x}}{\sqrt[]{2}}\right)+C$

It is given that the curve passes through

$\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt[]{2}}\right)$

$\frac{3}{\sqrt[]{2}}{e}^{\alpha }-\frac{3}{\sqrt[]{2}}=e^{\alpha }+\frac{9}{2}$

$e^{\alpha }=\frac{\frac{9}{2}+\frac{3}{\sqrt[]{2}}}{\frac{3}{\sqrt[]{2}}-1}=\frac{3}{\sqrt[]{2}}\left(\frac{3+\sqrt[]{2}}{3-\sqrt[]{2}}\right)$

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$  

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$  calculate C.

$\frac{dy}{dx}-y=x,$  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

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

Slope of any point P (x, y) to y = f (x) is $\frac{dy}{dx}=-k\frac{y}{x}$

$\Rightarrow \frac{dy}{y}+k\frac{dx}{x}=0$

Solving the equation the curve is xky = c

It passes (1, 2) c = 2 xky = 2 again it passes (8, 1) 8k = 2 k = $\frac{1}{3}$

$\therefore$ the equation of curve is $x^{1/3}$ y = 2 …… (i)

$\therefore \left|y\left(\frac{1}{8}\right)\right|=\left|\frac{2}{{\left(\frac{1}{8}\right)}^{1/3}}\right|=4$

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

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 =  $\frac{3}{2}$

If y = y(x), $x\in \left(0,\frac{\pi }{2}\right)$ $\frac{}{}$ be the solution curve of the different equation

$\frac{}{}$ $\therefore \frac{dy}{dx}+\left(8+4cot2x\right)y=\frac{2e^{-4x}}{si{n}^{2}2x}\left(2sinx+cos2x\right)$  which is a linear different equation

I.F = $e^{{\displaystyle \int \left(8+4cot2x\right)dx}}=e^{8x+2cos\left(sin2x\right)}=e^{8x}.si{n}^{2}2x$ ${}^{{\displaystyle }}{}^{}{}^{}{}^{}$

Given, $y\left(\frac{\pi }{4}\right)=e^{-\pi }\Rightarrow c=0$ $\frac{}{}{}^{}$

$\therefore y=\frac{e^{-4x}}{sin2x}$

$\therefore y\left(\frac{\pi }{6}\right)=\frac{e^{-4\frac{\pi }{6}}}{sin\left(2.\frac{\pi }{6}\right)}=\frac{2}{\sqrt[]{3}}{e}^{\frac{2\pi }{3}}$

  $\frac{dy}{dx}=\frac{x+y-2}{x-y}$  

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

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