Ok one more ODE: u'=(x(e^x))-u

[Schmoozer]

Homework Statement

"Solve the following ODE's:"
"u'+u=x(e^x)"

Homework Equations

N/A

The Attempt at a Solution

u'=(x(e^x))-u

u'=xe^x => u=(xe^x)/ln(xe)
u'=-u => u=e^-x

I'm unsure how to combine these two answers.

 

[Defennder]

u'=xe^x => u=(xe^x)/ln(xe)
u'=-u => u=e^-x

I don't follow what you are trying to do here. Where did ln come from? And how did you get u' = -u ? And what is "combine" suppose to mean?

The original DE is itself already in the form of a easily recognisable type of DE.

 

[Schmoozer]

What I was trying to do was separate the xe^x and the (-u) and deal with them independently and then combine them after words.

Int(A+B) is the same as Int(A)+Int(B) right?

But if their is a better way to do that please let me know which method to try, my diffy eq is very rusty.

 

[Defennder]

You can't just integrate it like that. You have to make sure the variables are properly separated before you can apply the separation of variables technique. In its present form, the DE isn't separable. A large factor in solving DE's analytically (by hand rather than numerical methods) is learning how to recognise a certain type of DE when you see one. What have you learned about how to identify this type of DE?

 

[Schmoozer]

Well I've been through diffy eq, but that was about 4 years ago. The review I've gotten for this class is:
1st or 2nd order ODE
Linear or Non Linear
Homogeneous or Non Homogeneous

We've reviewed several methods for solving these:
Seperable Equations
Method of Integration by Factors
Method of Variation of Parameters

I suppose the biggest problem is I've lost the small grasp I had on diffy eq. I just need to get back in practice. This is our first homework assignment, I know there is more to come.

Thanks for your help everyone!

 

[Defennder]

I can tell you for one thing the equation isn't separable, so you have to try out another approach. The method you should use is already outline in those which you said your class reviewed.