 widdonod1t

2021-12-29

Solve the following IVP for the exact equation and find the interval of validity for the solution. Barbara Meeker

Expert

For any differential equation that is given to be exact, we must have a function f(x,y) such that:

The given differential equation can be written as:
$2xy-9{x}^{2}+\left(2y+{x}^{2}+1\right)\frac{dy}{dx}=0$
$\left(2xy-9{x}^{2}\right)dx+\left(2y+{x}^{2}+1\right)dy=0$ (i)
Comparing to the standard form of $M\left(x,y\right)dx+N\left(x.y\right)dy=0$ we get:
$M\left(x,y\right)=2xy-9{x}^{2}$
$\therefore f\left(x,y\right)=\int \left(2xy-9{x}^{2}\right)dx+{\varphi }_{1}\left(y\right)$
$=2y\left(\frac{{x}^{2}}{2}\right)-9\left(\frac{{x}^{3}}{3}\right)+{\varphi }_{1}\left(y\right)$
$={x}^{2}y-3{x}^{3}+{\varphi }_{1}\left(y\right)$...(2)
$N\left(x,y\right)=2y+{x}^{2}+1$
$\therefore f\left(x,y\right)=\int \left(2y+{x}^{2}+1\right)dy+{\varphi }_{2}\left(x\right)$
$=2\left(\frac{{y}^{2}}{2}\right)+{x}^{2}\left(y\right)+\left(y\right)+{\varphi }_{2}\left(x\right)$ reinosodairyshm

Expert

$\left[y\left(2x\right)dx+{x}^{2}\left(dy\right)\right]+\left(2y+1\right)dy-9{x}^{2}dx=0$
$d\left({x}^{2}y\right)+d\left({y}^{2}+y\right)-d\left(3{x}^{3}\right)=0$
$⇒d\left({x}^{2}y+{y}^{2}+y-3{x}^{2}\right)=0$
$⇒{x}^{2}y+{y}^{2}+y-3{x}^{2}=c$
$y\left(0\right)=-3⇒0+4-3+0=c=6⇒{x}^{2}y+{y}^{2}+y-3{x}^{2}=6$ karton

Expert

Since ${x}^{2}y$ appears on both sides, cancel it on the yside, which gives $g\left(y\right)={y}^{2}+y$. Add this to the leftside and the general solution is:
$H\left(x,y\right)={x}^{2}y-3{x}^{2}+{y}^{2}+y$

Do you have a similar question?