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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?

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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    @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.2012-11-20
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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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    Another thing is that, you have to proceed with an iterative process till you find a suitable starting solution.2012-11-20