Question

determine whether the given differential equation is exact. if it is exact, solve it. (((3x^2)y)+e^y)dx) + (x^3 +xe^y - 2y) dy = 0

Answer 1

\[\begin{align*}
\left(3x^2y+e^y\right)dx+\left(x^3+xe^y-2y\right)dy&=0\\\\
\underbrace{\left(3x^2y+e^y\right)}_{M(x,y)}+\underbrace{\left(x^3+xe^y-2y\right)}_{N(x,y)}\frac{dy}{dx}&=0
\end{align*}\]
The equation will be exact if
\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]
Check the partial derivatives:
\[\frac{\partial }{\partial y}\left[3x^2y+e^y\right]=3x^2+e^y\\
\frac{\partial }{\partial x}\left[x^3+xe^y-2y\right]=3x^2+e^y\]
Indeed, the equation is exact.

Answer 2

We'll want to find a solution of the form \(\Psi(x,y)=C\), and we the equation is
\[\color{red}{\frac{\partial \Psi(x,y)}{\partial x}}+\color{blue}{\frac{\partial \Psi(x,y)}{\partial y}}\frac{dy}{dx}=0\]
(a result from the chain rule). We can equate \(M(x,y)\) with the red part, and \(N(x,y)\) with the blue.

Since \(M\) is the partial derivative of \(\Psi\) with respect to \(x\), integrating with respect to \(x\) gives
\[\begin{align*}
\frac{\partial \Psi}{\partial x}&=M(x,y)\\\\
\Psi&=\int M(x,y)~dx\\\\
\Psi&=\int \left(3x^2y+e^y\right)~dx\\\\
\Psi&=x^3y+xe^y+f(y)&\text{then differentiate with respect to }y\\\\
\frac{\partial \Psi}{\partial y}&=x^3+xe^y+f'(y)\\\\
N(x,y)&=x^3+xe^y+f'(y)&\text{since }N\text{ is the partial of }\Psi\text{ w.r.t. }y\\\\
x^3+xe^y-2y&=x^3+xe^y+f'(y)\\\\
-2y&=f'(y)\\\\
\frac{df(y)}{dy}&=-2y\\\\
\int df(y)&=-2\int y~dy\\\\
f(y)&=-y^2+C\end{align*}\]
So the solution is
\[\Psi=x^3y+xe^y-y^2+C\]
or simply
\[x^3y+xe^y-y^2=C\]