Question

I am needing help with finding an integrating factor and solving a given equation. For problems such as:

y + (2xy - e^(-2y))y' = 0

Answer 1

Not a linear equation, so I assume you mean the integrating factor an (in)exact equation.
\[y+(2xy-e^{-2y})y'=0\]
Here, \(M(x,y)=y\) and \(N(x,y)=2xy-e^{-2y}\).
\[\frac{\partial M}{\partial y}=1~~~~~~~~~\frac{\partial N}{\partial x}=2y\]
Clearly not equal. The integrating factor is given by
\[\mu(y)={\large e}^{\displaystyle\int\frac{2y-1}{y}~dy}=e^{2y-\ln y}=\frac{e^{2y}}{y}\]
\[\frac{e^{2y}}{y}\cdot y+\frac{e^{2y}}{y}(2xy-e^{-2y})y'=0\\
e^{2y}+\left(2xe^{2y}-\frac{1}{y}\right)y'=0\]
Equation is now exact, but just to check, let \(M=e^{2y}\) and \(N=2xe^{2y}-\dfrac{1}{y}\). Then,
\[\frac{\partial M}{\partial y}=2e^{2y}~~~~~~~~~\frac{\partial N}{\partial x}=2e^{2y}\]

Answer 2

Thank you! That was exactly what I needed. Your answer will help me apply this to the remaining questions in the section!