aspifsGak5u

2021-12-27

Differential Equations:
Identifying homogenous or non homogenous functions.
when $x=0,y=3$
$xydx-\left(x+2\right)dy=0$

temnimam2

$xydx-\left(x+2\right)dy=0$
$xydx=\left(x+2\right)dy$
$\frac{x}{x+2}dx=\frac{dy}{4}$
$\left(\frac{x+2-2}{x+2}\right)dx=\frac{dy}{4}$
$\left(\left(1\right)-\left(\frac{2}{x+2}\right)dx=\frac{dy}{4}$
Integrating both sides $\int dx-\int \frac{2}{x+2}dx=\int \frac{dy}{4}$
$x-2\mathrm{ln}|x+2|=m|y|+c$
when $x=0,4=3\left\{\therefore \text{given}\right\}$
$0-2\mathrm{ln}\left(2\right)=\mathrm{ln}3+c⇒c=-2\mathrm{ln}\left(2\right)-\mathrm{ln}3$
$⇒c=-2\left(0.693\right)-1.098$
$⇒c=-1.386-1.098$
$⇒c=-0.484$
$x-2\mathrm{ln}|x+2|=\mathrm{ln}|4|-2.484$
$⇒\mathrm{ln}|4|=x-2\mathrm{ln}|x+2|+2.484$

Serita Dewitt

$xydx-\left(x+2\right)dy=0⇔xydx=\left(x+2\right)dy⇔\frac{x}{\left(x+2\right)}dx=\frac{1}{y}dy$
$\int \frac{x}{\left(x+2\right)}dx=\int \frac{1}{y}dy$
$\int \frac{x}{\left(x+2\right)}dx=\mathrm{ln}|y|+A$
Use partial fractions to solve the remaining integral.
$\frac{x}{\left(x+2\right)}=\frac{A}{1}+\frac{B}{\left(x+2\right)}$
$x=A\left(x+2\right)+B⇒A=1,2A+B=0⇔A=1,B=\left(-2\right)$
$\frac{x}{\left(x+2\right)}=1-\frac{2}{\left(x+2\right)}$
Evaluate the new integrals
$\mathrm{ln}|y|+A=\int dx-\int \frac{2}{\left(x+2\right)}dx$
$\mathrm{ln}|y|+A=x-2\mathrm{ln}|\left(x+2\right)|+B$
Simplify the equation and then take exponentials.
$\mathrm{ln}|y|+2\mathrm{ln}|\left(x+2\right)|=x+C$
$\mathrm{ln}|y|+\mathrm{ln}|{\left(x+2\right)}^{2}|=x+C$
$\mathrm{ln}|y{\left(x+2\right)}^{2}|=x+C$
$y{\left(x+2\right)}^{2}=D{e}^{x}$
Answer: $y=D\frac{{e}^{x}}{{\left(x+2\right)}^{2}}$

karton

$xydx-\left(x+2\right)dy=0$
$xy-\left(x+2\right)\frac{dy}{dx}=0$
$xy-\left(x+2\right){y}^{\prime }=0$
Solve $\frac{1}{y}{y}^{\prime }=\frac{x}{x+2}$
$\mathrm{ln}\left(y\right)=x+2-2\mathrm{ln}\left(x+2\right)+{c}_{1}$
Isolate y: $y=\frac{{e}^{x+2+{c}_{1}}}{{x}^{2}+4x+4}$
Answer: $y=\frac{{e}^{x+2+{c}_{1}}}{{x}^{2}+4x+4}$

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