1. Jan 21, 2017

### Ric-Veda

1. The problem statement, all variables and given/known data
y''-16y=2e^4x. Find general solution

2. Relevant equations

3. The attempt at a solution
I have the homogenous equation which is c1e^-4x+c2e^4x, but I'm trying to find the particular solution for 2e^4x. I did yp=ae^4x, yp'=4ae^4x, yp''=16ae^4x, then plugged it into the equation, then got 0=2e^4x. What am I doing wrong. And I don't understand how 2e^4x becomes ae^4x

2. Jan 21, 2017

### Ray Vickson

Use the method of undermined coefficients; see, eg.,
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
or
http://www.math.psu.edu/tseng/class/Math251/Notes-2nd order ODE pt2.pdf

3. Jan 21, 2017

### LCKurtz

Since $e^{4x}$ is a solution to the homogeneous equation, of course trying $y_p = ae^{4x}$ is going to give you $0$. Try $y_p = Axe^{4x}$.

4. Jan 21, 2017

### Ric-Veda

But I need to know why instead of using yp=ae^4x, you have to use yp=axe^4x (sorry, the template to write the equation like you did is very complicated for me)

5. Jan 21, 2017

### Ric-Veda

My professor did not go in dept. I just know know:
if a constant: yp=A

if x: yp=Ax+B

if x^2: yp=Ax^2+Bx+C

if cos(x) or sin(x): yp=Acos(x)+Asin(x)

if e^x: yp=Ae^x

Or something like that???

6. Jan 21, 2017

### LCKurtz

7. Jan 21, 2017

### ehild

These are valid only, if the the right side is not solution of the homogeneous equation. If it is, you have to include the factor x, in order that you do not get zero on the right side when you substitute the particular solution.

Last edited: Jan 21, 2017
8. Jan 21, 2017

### Ray Vickson

Have you looked at the links I supplied in post #2?