# For David Shabazi (-----) Solve the differential equation (-----) Y' = Xe^- sin(x) - y cos(x)?

I'm not sure what the value of the power is. But I'll solve it the way I think it's written.

y = xe^(-sinx) - ycosx

y' = (x)[-cosx * e^(-sinx)] + e^(-sinx) - [(y)(-sinx) + cosx]

= -x*cosx*e^(-sinx) + e^(-sinx) + y*sinx - cosx

This is technically the answer, but it could be simplified further, which will result as the following:

y' = [e^(-sinx)](x*cosx + 1) + y*sinx - cosx

thanked the writer.
John McCann commented
Uh, no.

This is a first order linear differential equation with an integrating factor.

y = [(1/2)X^2 + C]e^(sinX)
=========================The answer ( reads....y equals 1/2 x squared + some constant times e to the siin of x ) You work on that a while and I will show you how to do this. Many scientific endeavors are reduced/explained by differential equations.

Looks like you tried implicit differentiation with this monster?!?
David Shabazi commented
I haven't learned this yet. My Calculus AB class started recently, so we haven't gotten to the chapter that covers this.
John McCann commented
OK.

I will pitch you softer balls from now on.

I was so going to answer this my my name isn't David shasomething or other ...HI " Tom " :p

2 People thanked the writer.
HappyTo BeHereTo commented
Don't blush Tom! Having people like you here keeps the site far more interesting and informative.

I'll demonstrate various ways for students to view the problem. You get to the heart of the matter quickly and concisely. You know more than I've forgotten.

@ Jaimie: Will he blush?
John McCann commented
This was back when Kos told me I should put some calculus questions on here. David is long gone and was one of perhaps three people on here who could handle intermediate mathematics.

One of those three people is not that smarty pants Tom Jackson!

You did notice this answer is old, didn't you?
HappyTo BeHereTo commented
I did. When Jaime played I couldn't resist the pun. ;) 