# (2xy - 3x^{2})dx + (x^{2} + y)dy = 0

$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
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Stella Calderon
The given differential equation is
$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
Comparing it with $Mdx+Mdy=0$

Here, $\frac{dM}{dy}=\frac{dN}{dx}$
Then, it is exact differential equation, so solution is
$\int Mdx+\int N$ (term not contining x) $dy=c$
$\int \left(2x-3{x}^{2}\right)dx+\int ydy=c$
$y\cdot \frac{2{x}^{2}}{2}-\frac{3{x}^{3}}{3}+\frac{{y}^{2}}{2}=c$
${x}^{2}y-{x}^{3}+\frac{{y}^{2}}{2}=c$
This is solution of given DE.
###### Not exactly what you’re looking for?
Charles Benedict
Assuming $\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$
Then $2xy+\left({x}^{2}+y\right)\frac{dy}{dx}=3{x}^{2}$ (1)
Notice that $d\frac{{x}^{2}y+\frac{{y}^{2}}{2}}{dx}$ is:
${x}^{2}\frac{dy}{dx}+2xy+2\frac{y}{2}\frac{dy}{dx}$
$=2xy+\left({x}^{2}+y\right)\frac{dy}{dx}$ which is left hand side of (1)
So integrating (1) gives
${x}^{2}y+\frac{{y}^{2}}{2}={x}^{3}+k$
So ${y}^{2}+2{x}^{2}y-2{x}^{3}+c=0$ is the solution
###### Not exactly what you’re looking for?
RizerMix

Let $M=\left(2xy-3{x}^{2}\right)$ so that $DM/Dy=2x$
Let $N=:\left({x}^{2}+y\right)$ so that $DN/Dx=2x$
since $DM/Dy=DN/Dx$ here D is to be read as delta.
The given Equation
$\left(2xy-3{x}^{2}\right)dx+\left({x}^{2}+y\right)dy=0$ is exact
Integrating $y{x}^{2}x-6x+1/2{y}^{2}=0$
or, $2y{x}^{2}x-12x+{y}^{2}=0$