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\[\Large y' + 2y = e ^{-x}\]

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))?

i didnt get the right answer from the formula