# First Order Linear Equations: Am I doing it right?

• Nov 30th 2009, 10:33 AM
TheTaoOfBill
First Order Linear Equations: Am I doing it right?
Just seeing if I understand this correctly.

Here is the problem.

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

I rewrote the problem to this form:
$dy/dx + ((2e^x)/e^x)y = 1/e^x$

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

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

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

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. $y(0) = 1$

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 $(e^x + c)/e^(2x)$ or is there more yet to do. Or did I just flat out do it wrong? Can someone help me!
• Nov 30th 2009, 10:54 AM
Hello TheTaoOfBill

Welcome to Math Help Forum!
Quote:

Originally Posted by TheTaoOfBill
Just seeing if I understand this correctly.

Here is the problem.

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

I rewrote the problem to this form:
$dy/dx + ((2e^x)/e^x)y = 1/e^x$

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

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

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

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. $y(0) = 1$

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 $(e^x + c)/e^(2x)$ 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:

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

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

You can easily check that the solution
$y = \frac{e^x+c}{e^{2x}}=e^{-x}+ce^{-2x}$
is correct, by differentiating:
$\frac{dy}{dx}=-e^{-x}-2ce^{-2x}$
and then substituting into the original differential equation.