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Find a particular solution for the differential equation by the method of undetermined coefficients. $$2y'' - 16y' + 32y = -e^{4x}$$ Also, find the general solution of this equation.

The steps I took to solve this problem,

Find the auxiliary equation which is $2m^2-16m+32=0$ for which the roots are $m_1=4$ and $m_2=4$ so $m=4$ of multiplicity 2.

Solve for a general equation of $y_h(x) = C_1e^{4x}x + C_2e^{4x}$

When I try to find a particular solution by taking the derivates of the right hand side, I get \begin{align} y_p &= Ae^{4x}\\ y_p' &= 4Ae^{4x}\\ y_p'' &= 16Ae^{4x} \end{align} Substituting these values into the left hand side results in $0 = -e^{4x}$ which is not possible. Can someone identify what I am missing?

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2 Answers 2

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Your problem is $e^{4x}$ is part of your homogeneous solution, which explains why you get $0$ when you try $Ae^{4x}$ as a particular solution.

I would try $Ax^2e^{4x}$ as a particular solution.

Just to show it works, let me show you another way to solve the problem. Let

$$z=y'-4y,z'=y''-4y'$$ $$y''-8y'+16y=-\frac12e^{4x}$$ $$(y'-4y)'-4(y'-4y)=z'-4z=-\frac12e^{4x}$$ $$e^{-4x}z'-4e^{-4x}z=(e^{-4x}z)'=-\frac12$$ $$e^{-4x}z=-\frac12x+k_1,z=-\frac12xe^{4x}+k_1e^{4x}$$ $$y'-4y=-\frac12xe^{4x}+k_1e^{4x}$$ $$e^{-4x}y'-4e^{-4x}y=(e^{-4x}y)'=-\frac12x+k_1$$ $$e^{-4x}y=-\frac14x^2+k_1x+k_2,y=-\frac14x^2e^{4x}+k_1xe^{4x}+k_2e^{4x}$$

So our particular solution turns out to be $-\frac14x^2e^{4x}$

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Why would we do this? –  Mike Beta Nov 20 '12 at 6:09
    
We start with a different starting point because a double root does not give us the two Linearly Independent solutions we are looking for. –  Aseem Dua Nov 20 '12 at 6:12
    
@MikeBeta Often in these kinds of problems, the solution is to multiply by $x$ when one particular solution doesn't work. $xe^{4x}$ is also part of the homogeneous solution, so multiply by $x$ again. If you want to try solving this problem a different way, try making the substitution $z=y'-4y$. It might help you understand. –  Mike Nov 20 '12 at 6:16

In case of a double root, the method fails, we need to start over with a different assumption of the solution. This time though, we approach the problem in a more informed manner.

hint: when we solve a simple homogeneous system, when we have a double root, to obtain a second LI solution we simply multiply root by x.

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This relates to the multiplicity of the root? Does the same happen for when the root is m = 4, 4, 4? –  Mike Beta Nov 20 '12 at 6:10
    
You start with an "anticipated" solution based on you qualitative understanding of the system. If it doesn't satisfy your requirements, you make an understanding and proceed to a next best possible attempt. In this case multiplicity of roots gives an understanding that the system will behave differently than "anticipated". If you are working with a higher order differential eq than 2, according to this logic, i feel yes you may have to change the initial attempt in case of a triple root. –  Aseem Dua Nov 20 '12 at 6:19
    
Another thing is that, you have to proceed with an iterative process till you find a suitable starting solution. –  Aseem Dua Nov 20 '12 at 6:26

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