Hello TheTaoOfBill

Welcome to Math Help Forum! Originally Posted by

**TheTaoOfBill** Just seeing if I understand this correctly.

Here is the problem.

$\displaystyle e^x(dy/dx) + (2e^x)y = 1$

I rewrote the problem to this form:

$\displaystyle dy/dx + ((2e^x)/e^x)y = 1/e^x$

then I found $\displaystyle e^(int((2e^x)/e^x)$ and got $\displaystyle e^(2x)$

Then I found the$\displaystyle int(e^(2x) * 1/e^x) $and got$\displaystyle e^x$

So I wrote the general formula as $\displaystyle (e^x + c)/e^(2x)$

At this point I wasn't sure if I was finished. I read a tutorial online that said I take it further to find C but they used an equation that was given at the start. $\displaystyle y(0) = 1$

in which case I would input 0 to get C+1 = 1 so C = 0

But that y(0) = 1 was never given.

So am I finished then? Is the answer just the general formula of$\displaystyle (e^x + c)/e^(2x)$ or is there more yet to do. Or did I just flat out do it wrong? Can someone help me!

Your answer is correct, but you could have simplified the equation to get:

$\displaystyle \frac{dy}{dx}+2y=e^{-x}$

before using the integrating factor $\displaystyle e^{\int2dx}=e^{2x}$

You can easily check that the solution

$\displaystyle y = \frac{e^x+c}{e^{2x}}=e^{-x}+ce^{-2x}$

is correct, by differentiating:

$\displaystyle \frac{dy}{dx}=-e^{-x}-2ce^{-2x}$

and then substituting into the original differential equation.

Grandad