Question

truing to do this differential equation:

Answer 1

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

Answer 2

but i can't get past this\[\Large \frac{ dy }{ dx } + 2y = e ^{-x}\]

Answer 3

for a lde\[dy/dx+p(x)y=q(x)\] solution is given as \[y=\int\limits_{?}^{?} q. e ^pxdx\]

Answer 4

sorry its e^pdx got an extra x by mistake

Answer 5

akash just wrote what you need to do:
y = Integral(q*e^(p(x)dx) wait is it e^(px) ir e^(-p(x))?

Answer 6

it is e^p(x).... @remnant did u try using the formula

Answer 7

i didnt get the right answer from the formula