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Find second order linear homogeneous ODE with constant coefficients if its fundamental set of solutions is {$e^{3t},te^{3t}$}.

Attempt: Had this question in my midterm. So, since the fundamental set of solutions is $$y=y_1+y_2=c_1e^{3t}+c_2te^{3t}$$ the characteristic equation of the second order ODE has only one root. I don't know what to do next. Help please.

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$(m-3)^2=0$.${}$ – David Mitra Feb 15 '13 at 19:14
up vote 4 down vote accepted

According to the theory of ODE with constant coefficients, the number of elements of a fundamental set of family of solutions coincides with the order of the ODE. So, your ODE here is second order. Moreover, if we consider the ODE with its standard form as $$ay''+by'+cy=0$$, so its associated auxiliary equation is $$am^2+bm+c=0.$$

Here you have $$m=3$$ of twice order, so the quadratic equation above is a complete square like $(m-3)^2=0$ and we know that in these cases we can find another solution, for example, by using the reduction method. This latter method gave us $t\exp(3t)$. Hence: $$(m-3)^2\longrightarrow m^2-6m+9=0\rightarrow y''-6y'+9=0$$

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very nice solution and probably the one their prof was looking for, +1. Regards – Amzoti Feb 15 '13 at 19:44
Nice, Babak! good to see you 'round! +1 – amWhy Feb 16 '13 at 0:08

The eigenvalues and eigenvectors for the coefficient matrix $A$ in the linear homogeneous system:

$Y'= AY$ are $\lambda_{1} = 3$ with $v_1 =< a; b >$ and $\lambda_2 = 3$ with $v2 =< c; d >$

The fundamental form of the solution is:

$$ Y = c_1 e^{3t}v_1 + c_2t e^{3t}v_2$$

Take the second derivative, $Y''$ for the DEQ.

Your original system will be of the form:

$$y'' - 6y' + 9 = 0$$

to give you the double eigenvalue $\lambda_{1,2} = 3$. You can actually solve this to find the corresponding eigenvectors.


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+1. You did it via matrices! – Babak S. Feb 15 '13 at 19:43

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