hunterofdeath63

2021-12-30

Find the solution of the following differential equations:
Homogeneous Differential Equations
$Зх{у}^{2}dx+\left(3{x}^{2}у+{у}^{3}\right)dy=0;y\left(1\right)=2$
$\left({x}^{2}-9{y}^{2}\right)dx+2xydy=0;y\left(2\right)=-1$

Matthew Rodriguez

Expert

The given homogenous differential equation
$3x{y}^{2}dx+\left(3{x}^{2}y+{y}^{3}\right)dy=0$ (1)
$y\left(1\right)=2$ (2)
First we check the Exactness of the given Differential equation,
$Mdx+Ndy=0$
$\therefore M=3x{y}^{2},N=3{x}^{2}y+{y}^{3}$
Then, $\frac{\partial M}{\partial y}=6xy,\frac{\partial N}{\partial x}=6xy$
$\therefore \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
So, the given differential equation is exact.
Hence the solution is given by.
$\int Mdx+\int \left(\text{terms in N not containg x}\right)dy=c$
$\int 3x{y}^{2}dx+\int {y}^{3}dy=c$
$3{y}^{2}\cdot \frac{{x}^{2}}{2}+\frac{{y}^{4}}{4}=c$ (3)
$\frac{3}{2}{x}^{2}{y}^{2}+\frac{{y}^{4}}{4}=c$
using (2)
Put $x=1,y=2$
$3{\left(1\right)}^{2}{\left(2\right)}^{2}+\frac{{\left(2\right)}^{4}}{4}=c$
$6+4=c⇒c=10$
(3) becomes
$\frac{3}{2}{x}^{2}{y}^{2}+\frac{{y}^{4}}{4}=10$
$3{x}^{2}{y}^{2}+{y}^{4}=40$ This is the required solution.

einfachmoipf

Expert

$\left({x}^{2}-9{y}^{2}\right)dx+2xydy=0$
$y\left(2\right)=-1$
$\left({x}^{2}-9{y}^{2}\right)dx=-2xydy$
$\frac{dy}{dx}=\frac{-\left({x}^{2}-9{y}^{2}\right)}{2xy}$
$y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$v+x\frac{dv}{dx}=\frac{-\left({x}^{2}-9{v}^{2}{x}^{2}\right)}{2×x×vx}$
$v+x\frac{dv}{dx}=\frac{-{x}^{2}\left(1-9{v}^{2}\right)}{2{x}^{2}×v}=\frac{-\left(1-9{v}^{2}\right)}{2v}$
$x\frac{dv}{dx}=\frac{-1+9{v}^{2}-2{v}^{2}}{2v}=\frac{-1+7{v}^{2}}{2v}$
$\int \frac{2v}{7{v}^{2}-1}dv=\int \frac{dx}{x}$
Let $7{v}^{2}-1=f$
$14\cdot b\cdot dv=dt⇒2vdv=\frac{dt}{7}$
$\frac{1}{7}\int \frac{dt}{t}=\int \frac{dt}{x}$
$\frac{1}{7}\mathrm{ln}\left(t\right)=\mathrm{ln}\left(x\right)+\mathrm{ln}c$
${\mathrm{ln}\left(7{v}^{2}-1\right)}^{\frac{1}{7}}=\mathrm{ln}\left(x\cdot c\right)$

Vasquez

Expert