For the initial-value problem y' + 3y = 2xe^(-5x), how to do I find the solution if y(0) = C? I know that it is not separable, so how do I go on from there? Integrate?

Question

For the initial-value problem y' + 3y = 2xe^(-5x), how to do I find the solution if y(0) = C? I know that it is not separable, so how do I go  on from there? Integrate?

turingtest · Answer

This is a linear DE. Use an integrating factor.

anonymous · Answer

I am not sure what you mean by integrating factor...I did a double integral, and that seemed wrong...

anonymous · Answer

I guess I don't know what to do in order to get to the point where I plug in my initial value, since it doesn't solve as nicely as separable ones do.

turingtest · Answer

It is an integrating factor of a DE of the form$$y'+p(x)y=q(x)$$is$$e^{\int p(x)dx}$$multiply the equation by that and you will be able to integrate. If you have never seen this technique before you may want to read up on it. http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

jamesj · Answer

Right. Integrating factors is normally the method you learn after separation of variables. You might also find this lecture helpful: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

turingtest · Answer

after integrating you can use your initial value to find the value of C,
but in this case they just tell you that y(0)=C, so I don't see the point of that...

anonymous · Answer

Ooooh---so we haven't gotten to this in class. Thanks everyone for the links---I had reviewed the Paul's online one earlier, but it seems I need to do some rereading...
Yeah, I didn't really understand the initial condition's purpose either

turingtest · Answer

The MIT link James posted is better than Paul, try to understand that.

jamesj · Answer

The point of y(0) = c is that the constant of integration isn't the same as c.

anonymous · Answer

oh, okay. But, so you mean what ever I may get for, say, C_1 isn't going to be the same as C? Although in this case I don't think I would have numbers---but in a general case?

jamesj · Answer

The point of asking you to find the solution for y(0) = c is just you to work through how to find a particular solution; that is, solve the Initial Value Problem; vs. just finding the general solution.

Anyway, when you solve this equation you'll see how this works.

anonymous · Answer

Thanks everyone!

jamesj · Answer

There's about 15 minutes in that video lecture at the beginning dealing with certain models.  Don't let that throw you.  He does get to the general method after them.  

But in any case, those models and their variants come up all the time in ODE modeling, and they're worth learning in their own right.