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, 03:31 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, 03:32 PMAckbeet
What ideas have you had so far?

- Jan 8th 2011, 03:38 PMdwsmith
- Jan 8th 2011, 03:49 PMProve It
For part 2, you're told $\displaystyle \displaystyle y = e^{3x}$ is a solution.

So $\displaystyle \displaystyle y' = 3e^{3x}$ and $\displaystyle \displaystyle y'' = 9e^{3x}$.

Substitute into the DE to get $\displaystyle \displaystyle 9e^{3x} + 3a\,e^{3x} + b\,e^{3x} = 0$.

Now do the same with the other solution $\displaystyle \displaystyle y = x\,e^{3x}$ to get another equation that involves $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$. Then you can solve the two equations simultaneously for $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$. - Jan 8th 2011, 04:00 PMdwsmith
- Jan 8th 2011, 04:02 PMhejmh
Yes. thanks

- Jan 9th 2011, 07:19 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 $\displaystyle u(x)= y'(x)= e^{-4x+ c}= Ce^{-4x}$ where $\displaystyle C= e^{c}$. Now find y by integrating $\displaystyle y'= Ce^{-4x}$: $\displaystyle y(x)= \frac{C}{-4}e^{4x}+ B$ which we can also write as $\displaystyle y(x)= Ae^{-4x}+ B$ where now $\displaystyle A= \frac{C}{-4}= \frac{e^c}{-4}$.