Here's the question you clicked on:
remnant
truing to do this differential equation:
\[\Large y' + 2y = e ^{-x}\]
but i can't get past this\[\Large \frac{ dy }{ dx } + 2y = e ^{-x}\]
for a lde\[dy/dx+p(x)y=q(x)\] solution is given as \[y=\int\limits_{?}^{?} q. e ^pxdx\]
sorry its e^pdx got an extra x by mistake
akash just wrote what you need to do: y = Integral(q*e^(p(x)dx) wait is it e^(px) ir e^(-p(x))?
it is e^p(x).... @remnant did u try using the formula
i didnt get the right answer from the formula