The general solution for the differential equation involving \( ye^x \frac{dy}{dx} = e^{-y} + e^{-2x-y} \) can be simplified to \( ye^y dy = \frac{1 + e^{-2x}}{e^x} dx \). This equation is not homogeneous, and integration by parts is recommended, using \( u = y \) and \( dv = e^y dy \) for the left side. The right side can be rewritten as \( e^{-x} + e^{-3x} \) for easier integration.
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It is very difficult to understand what you mean. Do you mean that you are asked to find the general solution? And is "e^-2x- y" supposed to be "e^(-2x-y)" or "e^(-2x) - y"? Guessing at what you mean:intenzxboi said:the general solution for ye^x dy/dx = [e^-y ] + [ e^ -2x-y ]is:
so i tried to see if this was a homogenous equation but it is not.
next i tried to simplify and got:
y e^y dy= (1+e^-2x) dx / e^x