Chesley

2020-12-29

Solve differential equation $2xy-9{x}^{2}+\left(2y+{x}^{2}+1\right)dy/dx=0$, y(0)= -3

2abehn

$\left(2xy-9{x}^{2}\right)dx+\left(2y+{x}^{2}+1\right)dy=0$
Comparing with M dx + N dy = 0 we get
$M=2xy-9{x}^{2}$, $N=2y+{x}^{2}+1$
$\left(dM\right)/dy=2x-0=2x$
$\left(dN\right)/dx=0+2x+0=2x$
$\left(dM\right)/dy=\left(dN\right)/dx$
int Mdx+int (terms of N not containing x) dy=C
$=>\int \left(2xy-9{x}^{2}\right)dx+\int \left(2y+1\right)dy=C$
$=>\left(2y{x}^{2}/2\right)-9{x}^{3}/3+2{y}^{2}/2+y=C$
$\int {x}^{2}y-3{x}^{3}+{y}^{2}+y=C$
$\int {y}^{2}+\left({x}^{2}+1\right)y-3{x}^{3}=C$ (1)
Given y(0) = −3
substitute x=0 and y= -3 in (1) we get
$\left(-3{\right)}^{2}+\left({0}^{2}+1\right)\left(-3\right)-3\left({0}^{3}\right)=C$
9−3=C int C=6
substitute C=6 in (1) we get
${y}^{2}+\left({x}^{2}+1\right)y-3{x}^{3}=6$
$\int {y}^{2}+\left({x}^{2}+1\right)y-3{x}^{3}-6=0$
$2xy-9{x}^{2}+\left(2y+{x}^{2}+1\right)dy/dx=0$
$y\left(0\right)=-3is{y}^{2}+\left({x}^{2}+1\right)y$
$-3{x}^{3}-6=0$

Jeffrey Jordon

Answer is given below (on video)

Jeffrey Jordon