Cheexorgeny

2021-12-31

Determine which of the following differential equations are homogeneous, then solve the general solution. If initial condition is given, find the particular solution:
$xydx-{\left(x+2y\right)}^{2}dy=0$

Jim Hunt

$xydx-{\left(x+2y\right)}^{2}dy=0$
$xydx={\left(x+2y\right)}^{2}dy$
$\frac{dx}{dy}=\frac{{x}^{2}+4{y}^{2}+4xy}{xy}$
$\frac{dx}{dy}=\left(\frac{x}{y}\right)+4\left(\frac{y}{x}\right)+4$
It is a Homogeneous differential equation.
Let $\frac{x}{y}=v⇒x=vy$
So, $\frac{dx}{dy}=v+y\frac{dv}{dy}$
$\therefore v+y\frac{dv}{dy}=v+\frac{4}{v}+4$
$y\frac{dv}{dy}=4\left(\frac{v+1}{v}\right)$
$\frac{vdv}{v+1}=4\frac{dy}{y}$
It becomes variable separable equation. So, now integrate both sides.
$\int \frac{vdv}{v+1}=4\int \frac{dy}{y}$
$\int \left(\frac{v+1-1}{v+1}\right)dv=4\int \frac{dy}{y}$
$\int \left(1-\frac{1}{v+1}\right)dv=4\mathrm{ln}|y|+\mathrm{ln}c$
$v-\mathrm{ln}|v+1|=\mathrm{ln}\left({y}^{4}c\right)$
$\therefore \frac{x}{y}+\mathrm{ln}|\frac{x}{y}+1|=\mathrm{ln}|{y}^{4}c|$
$\frac{x}{y}=\mathrm{ln}|{y}^{4}c|-\mathrm{ln}|\frac{x+y}{y}|$

Serita Dewitt

$xydx-{\left(x+2y\right)}^{2}dy=0$
$xydx={\left(x+2y\right)}^{2}dy$
$\frac{dx}{dy}=\frac{{\left(x+2y\right)}^{2}}{xy}$
$\frac{dx}{dy}=\frac{{x}^{2}+4{y}^{2}+4xy}{xy}$
$\frac{dx}{dy}=\frac{x}{y}+4\frac{y}{x}+4$.
Substituting; $x=yv$
Diff. w.r.t. y
$\frac{dx}{dy}=v+y\frac{dv}{dy}$
$v+y\frac{dv}{dy}=v+4\frac{1}{v}+4$
$\mathrm{¬}\left\{v\right\}+y\frac{dv}{dy}=\mathrm{¬}\left\{v\right\}+4\left(\frac{1}{v}+1\right)$
$\frac{y}{4}\frac{dv}{dy}=\frac{1+v}{v}$
Integrating $\int \frac{v}{v+1}dv=4\int \frac{1}{y}dy$
$\int \frac{v+1-1}{v+1}dv=4\int \frac{1}{y}dy$
$v-\mathrm{ln}|v+1|=4\mathrm{log}y+c$
applying $v=\frac{x}{y}$
$\frac{x}{y}-\mathrm{ln}|\frac{x}{y}+1|=4\mathrm{log}y+c$

karton

$\begin{array}{}xydx-\left(x+2y{\right)}^{2}dy=0\\ xydx=\left(x+2y{\right)}^{2}dy\\ \frac{dx}{dy}=\frac{{x}^{2}+4{y}^{2}+4xy}{xy}\\ \frac{dx}{dy}=\left(\frac{x}{y}\right)+4\left(\frac{y}{x}\right)+4\\ \frac{x}{y}=v⇒x=vy\\ \frac{dx}{dy}=v+y\frac{dv}{dy}\\ \therefore \text{⧸}v+y\frac{dv}{dy}=\text{⧸}v+\frac{4}{v}+4\\ y\frac{dv}{dy}=4\left(\frac{v+1}{v}\right)\\ \frac{vdv}{v+1}=4\frac{dy}{y}\\ \int \frac{vdv}{v+1}=4\int \frac{dy}{y}\\ \int \left(\frac{v+1-1}{v+1}\right)dv=4\int \frac{dy}{y}\\ \int \left(1-\frac{1}{v+1}\right)dv=4\mathrm{ln}|y|+\mathrm{ln}c\\ v-\mathrm{ln}|v+1|=\mathrm{ln}\left({y}^{4}c\right)\\ \therefore \frac{x}{y}+\mathrm{ln}|\frac{x}{y}+1|=\mathrm{ln}|{y}^{4}c|\\ \frac{x}{y}=\mathrm{ln}|{y}^{4}c|-\mathrm{ln}|\frac{x+y}{y}|\\ \frac{x}{y}=\mathrm{ln}|\frac{\left({y}^{4}c\right)}{\left(\frac{x+y}{y}\right)}|\\ x=y\mathrm{ln}|\frac{c{y}^{5}}{x+y}|\end{array}$

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