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Im trying to find the particular solution to: y'' - 3y' + 2y = 2e^x

I already have the homogenous solution so this is not my problem.

Assuming that y_p = Ae^x \to {y'}_p = Ae^x \to {y''}_p = Ae^x we have:

$Ae^x - 3Ae^x + 2Ae^x = 2e^x$ $0 \cdot Ae^x = 2e^x$ $0 = 2$

Which is very bad. What am I doing wrong here?

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    Any differential equations textbook, when it discusses the method of undetermined coefficients, should explain what to do, even if it does not explain the theoretical reasons for doing that.2011-12-21

3 Answers 3

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The simplest thing in your case is probably to use this (easily justified) trick:

If $p(x)$ is a one variable polynomial with complex coefficients and $c$ a complex number, then the ODE $ p\left(\frac{d}{dx}\right)y=e^{cx} $ has a solution of the form $e^{cx}\,q(x)$ where $q(x)$ is a polynomial whose degree is the number of roots of $p(x)$ equal to $c$.

In you case, $p(x)=(x-1)(x-2)$, $c=1$, so the trick tells you to look for a solution of the form $ (ax+b)\ e^x, $ and you can clearly assume $b=0$.

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$y^{\prime \prime}−3y^{\prime}+2y= 2e^x$. Use operator D:

Let $D = d/dx$. So, $(D^2 - 3D + 2)y_p = 2e^x \quad \Rightarrow \quad (D-1)(D-2)y_p = 2e^x \quad \Rightarrow$

$y_p = \dfrac{1}{(D-1)(D-2)}\cdot 2e^x = 2e^x\dfrac{1}{(D+1-1)(D+1-2)}\cdot 1 = 2e^x\dfrac{1}{D(D-1)}\cdot 1 \quad \Rightarrow$

$y_p = 2e^x\dfrac{1}{D(0-1)\cdot 1} = -2e^x\dfrac{1}{D}\cdot 1 = -2xe^x$

Look this blog: http://fatosmatematicos.blogspot.com/2011/02/o-metodo-da-chave-invertida-para-achar.html

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This is too long for a comment, so I posted it as an answer. First solve for the homogeneous equation y'' - 3y' + 2y =0 by setting the right hand side to be zero. The auxiliary equation is $m^2-3m+2=0$, which has roots $m=2,1$. Therefore the solution for this homogeneous equation is $e^x$ and $e^{2x}$. Now we want to find a particular solution y'' - 3y' + 2y = 2e^x. Normally we set the particular solution to be $Ae^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with $x$ until no duplication occurs. Therefore, the particular solution is given by $Axe^{x}$.

Let me do another example: to solve y'' - 2y' + y = 2e^x. First solve the homogenous equation y'' - 2y' + y = 0. The auxiliary equation is $m^2-2m+1=0$ which has double roots $m=1$. Therefore, the solution for this homogeneous equation is $e^x$ and $xe^{x}$. Now if we want to find a particular solution y'' - 2y' + y = 2e^x. Normally we set the particular solution to be $Ae^{x}$. However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with $x$ and it becomes $Axe^x$, but it still duplicates with $xe^{x}$. Therefore, we mupltiply it by $x^2$, and the particular solution is given by $Ax^2e^{x}$.