Solving Basic Equation: y' + 2y = 4x - Tips and Tricks

[Icebreaker]

I've delayed my ODE and PDE classes as much as I could because I knew it would give me constant and continuous headaches. And I was right.

Can someone help me solve this basic equation:

y' + 2y = 4x

I got y = 4x, which makes no sense.

 

[HallsofIvy]

You're right, it makes no sense since if y= 4x, y'= 4 so y'+ 2y= 4+ 8x which looks nothing like 4x! How did you get "y= 4x"?

This is a linear, nonhomogeneous differential equation which I'm sure your textbook talks about. Do you know how to find an "integrating factor". (Every first order differential equation has an "integrating factor" that makes it very easy to solve. Unfortunately, for most equations finding the integrating factor is as hard as solving the equation. For linear equations, however, there is a specific formula that I'll bet is given in your textbook! Look under the heading "linear equations" in the first or second chapters.

 

[saltydog]

I've delayed my ODE and PDE classes as much as I could because I knew it would give me constant and continuous headaches. And I was right.

Can someone help me solve this basic equation:

y' + 2y = 4x

I got y = 4x, which makes no sense.

Well Icebreaker, whenever you have a first-order ODE like that, calculate the integrating factor and multiply both sides by it. In your case the integrating factor, designated by sigma is:

\sigma=e^{2x}

After multiplying both sides by that, the LHS becomes an exact differential and you're left with:

d\left(ye^{2x}\right)=4xe^{2x}

Integrating both sides:

ye^{2x}=\int 4xe^{2x}

I bet you can finish it (solving for y) and don't forget the constant of integration.

Feel like doin' 9 more?

 

[Icebreaker]

One down, 19 to go... Thanks for the help.