1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:

e^(3x) and xe^(3x)

Thanks for your help.

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- Jan 8th 2011, 04:31 PMhejmh2nd order DE's with constant coefficients.
1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:

e^(3x) and xe^(3x)

Thanks for your help. - Jan 8th 2011, 04:32 PMAckbeet
What ideas have you had so far?

- Jan 8th 2011, 04:38 PMdwsmith
- Jan 8th 2011, 04:49 PMProve It
For part 2, you're told is a solution.

So and .

Substitute into the DE to get .

Now do the same with the other solution to get another equation that involves and . Then you can solve the two equations simultaneously for and . - Jan 8th 2011, 05:00 PMdwsmith
- Jan 8th 2011, 05:02 PMhejmh
Yes. thanks

- Jan 9th 2011, 08:19 AMHallsofIvy
Here's another way to do this. Since there is no "y" (undifferentiated) let u= y' so the equation becomes u'+ 4u= 0. That is the same as du/dx= -4u which is a separable, first order differential equation. You can write it as du/u= -4 dx and integrate both sides: ln(u)= -4x+ c which leads to where . Now find y by integrating : which we can also write as where now .