Question

differential equation. ydx+(2xy-e^(-2y)dy=0 I know it's an exact equation but can't figure it out

Answer 1

\[y~dx+(2xy-e^{-2y})~dy=0\]
You have \(M(x,y)=y\) and \(N(x,y)=2xy-e^{-2y}\), and \(\dfrac{\partial M}{\partial y}=1\), \(\dfrac{\partial N}{\partial x}=2y\). So the equation is not exact.

Your integrating factor would be
\[\mu(y)=\exp\left(\int\frac{2y-1}{y}~dy\right)=\exp(2y-\ln y)=\frac{e^{2y}}{y}\]

Answer 2

unfortunately my professor is not good at explaining things. I don't even know if we learned what to do if an equation is not exact. Do i take the integrating factor, multiply both sides then take the integral?

Answer 3

An equation is not exact if \(M_y\not=N_x\), where \(M_y\) denotes the partial derivative of \(M\) with respect to \(y\), and similarly for \(N_x\) with respect to \(x\).

The integrating factor is given by either
\[\exp\left(\int\frac{N_x-M_y}{M}~dy\right)~~\text{or}~~\exp\left(\int\frac{M_y-N_x}{N}~dx\right)\]
Which you use depends on how complex the integral looks. Always go for the simpler one, meaning the one with only one variable. Sometimes the problem isn't as convenient as this one.

In this case,
\[\mu(y)=\frac{e^{2y}}{y}\]
You would then multiply both sides of the equation by \(\mu\):
\[y~dx+(2xy-e^{-2y})~dy=0~~\iff~~y+(2xy-e^{-2y})\frac{dy}{dx}=0\\
\mu y+\mu(2xy-e^{-2y})y'=0\\
e^{2y}+\left(2xe^{2y}-\frac{1}{y}\right)y'=0\]
Now you have \(M_y=2e^{2y}\) and \(N_x=2e^{2y}\), and hence an exact equation. Do you know how to solve equations that *are* exact?