Question

Differential Equation. e^(x)ydy/dx=e^(-y)+e^(-2x-y) I can't seem to figure this out.

Answer 1

Is it:
\[e^xy\frac{dy}{dx}=e^{-y}+e^{-2x-y}\]
?

Answer 2

Yes that is correct

Answer 3

\[\eqalign{
&e^xy\frac{dy}{dx}=e^{-y}+e^{-2x-y} \\
&e^xy\frac{dy}{dx}=e^{-y}+e^{-2x}e^{-y} \\
&e^xy\frac{dy}{dx}=e^{-y}(1+e^{-2x}) \\
&\frac{y}{e^{-y}}\frac{dy}{dx}=\frac{1+e^{-2x}}{e^x} \\
&y(e)^ydy=\frac{1+e^{-2x}}{e^x}dx \\
&y(e)^ydy=e^{-x}+e^{-3x}dx \\
&\int y(e)^ydy=\int e^{-x}+e^{-3x}dx \\
&\phantom{.......}LS=RS
}\]
Let's compute each side

Answer 4

\[LS=\int ye^ydy=\int u\phantom{.}dv\]
\[\eqalign{
&u=y\to du=(1) \\
&dv=e^y\to v=e^y \\
}\]
\[\eqalign{
LS=\int ye^ydy=\int u\phantom{.}dv&=uv-\int v\phantom{.}du \\
&=ye^y-\int e^y \\
&=ye^y-e^y
}\]

Answer 5

\[RS=\int e^{-x}+e^{-3x}dx=-e^{-x}-\frac{1}{3}e^{-3x}+c_1\]

Answer 6

Therefore, we obtain:
\[ye^y-e^y=-e^{-x}-\frac{1}{3}e^{-3x}+c_1\]

Answer 7

Thank you very much. I just had a problem finding what method to use.

Answer 8

You would try and separate the ODE :)