Find $M(x, y)$ for which $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation. Title says it all.
Find $M(x, y)$ so that $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.
I tried to use the $d M(x,y)/dy = d N(x,y)/dx$ formula but got stuck. I guess we must cancel the derivatives by applying an integral in both sides but I'm not sure how.
 A: hint
Your condition can be written as
$$M(x,y)dx+(3xy^2+20x^2y^3)dy=0=dU$$
with
$$\frac{\partial U}{\partial y}=3xy^2+20x^2y^3$$
so
$$U=xy^3+5x^2y^4+C(x)$$
and
$$M(x,y)=\frac{\partial U}{\partial x}=...$$
A: Just a hint:
Write $y'$ as $\frac{dy}{dx}$ till the ODE looks like
$$M(x,y)dx+(3xy^2+20x^2y^3)dy=0.$$
Now, find the possible function for $M(x,y)$ such that $M_y=N_x.$
A: I hope that this is right, if not please correct me.
If $M(x,y)$ is an exact equation, it's true that:
$\frac{dM(x,y)}{dy}=\frac{dN(x,y)}{dx}$
Then:
$\frac{dM(x,y)}{dy}= \frac{d(3xy^2+20x^2y^3)}{dx}$
$\frac{dM(x,y)}{dy}= 3y^2+40xy^3$
$dM(x,y)= (3y^2+40xy^3)dy$
$\int{dM(x,y)}= \int{(3y^2+40xy^3)dy}$
Therefore:
$M(x,y)= y^3+10xy^4+c$
A: $$M(x,y)dx+(3xy^2+20x^2y^3)dy=0$$
$$M(x,y)dx+xd(y^3)+5x^2d(y^4)=0$$
$$M_1(x,y)dx+xd(y^3)+M_2(x,y)dx+5x^2d(y^4)=0$$
$$(\color{red}{y^3dx}+xd(y^3))+(\color{red}{5y^4dx^2}+5x^2d(y^4))=0$$
$$d(y^3x)+5d(y^4x^2)=0$$
$$y^3x+5y^4x^2=C$$
There are many other solutions since you can write $M(x,y)$ as $M_1(x,y)dx+M_2(x,y)dx+f(x)dx$.
So that:
$$M(x,y)=y^3+10y^4x+f(x)$$
