Please help with this Differential equation problem

[Ric-Veda]

Homework Statement

y''-16y=2e^4x. Find general solution

Homework Equations

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

 

[Ray Vickson]

Homework Statement

y''-16y=2e^4x. Find general solution

Homework Equations

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

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%20order%20ODE%20pt2.pdf

 

[LCKurtz]

Homework Statement

y''-16y=2e^4x. Find general solution

Homework Equations

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

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

 

[Ric-Veda]

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

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)

 

[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?

 

[LCKurtz]

You have been suggested to look at the method of undetermined coefficients above. Another more organized way is to use the method of annihilators, which involves less guesswork. Lots of info on the internet. One source is
http://dankalman.net/AUhome/classes/classesF12/odes/assignments/annihilatormethod.pdf

 

[ehild]

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

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

if e^x: yp=Ae^x

Or something like that?

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.

 

[Ray Vickson]

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)

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