Tiffany Russell

2021-12-31

Find the solution of the following Differential Equations by Exact $\left(2y+xy\right)dx+2xdy=0,y\left(3\right)=\sqrt{2}$

esfloravaou

Expert

Given:
$\left(2x+xy\right)dx+2xdy=0,y\left(3\right)=\sqrt{2}$
this equation is form $M\left(x,y\right)dx+N\left(x,y\right)dy=0$
At this equation is exact if and only it
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Hence $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
Now we find the integrating factor to made this equation exact.
$My-Nx=2+x-2=x$
$\frac{My-Nx}{N}=\frac{x}{2x}=\frac{1}{2}$
Hence $I.F.={e}^{\int \frac{1}{2}dx}={e}^{\frac{x}{2}}$
$⇒{e}^{\frac{x}{2}}\left(2y+xy\right)dx+2{e}^{\frac{x}{2}}xdy=0$ (2)
$=2{e}^{2/2}y+x{e}^{x/2}y$
$N=2{e}^{\frac{x}{2}}x$
$\frac{\partial M}{\partial y}=2{e}^{\frac{x}{2}}+x{e}^{\frac{x}{2}}$
$\frac{\partial N}{\partial x}=2{e}^{\frac{x}{2}}y+2x\cdot {e}^{\frac{x}{2}}\cdot \frac{1}{2}=2{e}^{\frac{x}{2}}+x{e}^{\frac{x}{2}}$
$⇒\frac{\partial M}{\partial y}=\frac{\partial M}{\partial x}$
equation (2) is exact equation:
Now solution of equation

$\int \left(2y{e}^{\frac{x}{2}}+xy{e}^{\frac{x}{2}}\right)dx+\int dy=c$

eskalopit

Expert

$\left(2y+xy\right)dx+2xdy=0$
rearranging,
$\frac{dy}{dx}=-\frac{2y+xy}{2x}=-\frac{2+x}{2x}.y=-g\left(x\right).y$
$\frac{dy}{dx}+g\left(x\right).y=0$
Integrating Factor
$IF=\mathrm{exp}\left(\int g\left(x\right)dx\right)$
$\int g\left(x\right)dx=\int \frac{2+x}{2x}dx=\int \left\{\frac{1}{x}+\frac{1}{2}\right\}dx$
$\int g\left(x\right)dx=\mathrm{ln}\left(x\right)+\frac{x}{2}$
Hence,
$IF=\mathrm{exp}\left(\mathrm{ln}\left(x\right)+\frac{x}{2}\right)=x\cdot {e}^{\frac{x}{2}}$
The DE now becomes,
$d\left(IF\cdot y\right)=0$
$d\left(x\cdot {e}^{\frac{x}{2}}.y\right)=0$
Integrating,
$x\cdot {e}^{\frac{x}{2}}.y=const$
Answer: $xy.{e}^{\frac{x}{2}}=c$

karton

Expert