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Please help with this Differential equation problem

  • Thread starter Ric-Veda
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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
 

Ray Vickson

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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
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
 

LCKurtz

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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
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}##.
 
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)
 
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

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ehild

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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.

upload_2017-1-22_5-34-12.png
 
Last edited:

Ray Vickson

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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?
 

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