Consider the differential equation(adsbygoogle = window.adsbygoogle || []).push({});

[tex]t^2x''-tx'+x=0[/tex]

Given that [tex]x(t)=t[/tex] is a solution of the ordinary differential equation, find another linearly independent solution.

[tex]\textsc{SOLUTION}[/tex]

Since we know [tex]x(t) = t[/tex] is a solution, the general solution of the ODE will be of the form [tex]x(t) = Kt + Kty(t)[/tex], where [tex]y(t)[/tex] is some function linearly independent to [tex]x(t) = t[/tex]. This suggests that we utilize a solution of the form:

[tex]x(t) = ty(t)[/tex]

Take [tex]x(t) = ty(t)[/tex]. Then

[tex]x'(t) = y(t) + ty'(t)[/tex]

[tex]x''(t) = 2y'(t) + y''(t)[/tex]

Substitute this back into the original ODE we get

[tex]t^2\left(2y'(t) + y''(t)\right) - t\left(y(t) + ty'(t)\right) + ty(t) = 0[/tex]

[tex]t^2y''(t)+t^2y'(t) -ty(t) +ty(t) = 0[/tex]

[tex]t^2y''(t)+t^2y'(t) = 0[/tex]

[tex]t^2\left(y''(t)+y'(t)\right) = 0[/tex]

[tex]y''(t) + y'(t) = 0[/tex]

Integrating both sides with respect to [tex]t[/tex] we obtain

[tex]y'(t) + y(t) = C[/tex]

Rearranging

[tex]y'(t) = -y(t) + C[/tex]

This differential equation is separable

[tex]\frac{dy(t)}{dt} = -y(t) + C[/tex]

[tex]\frac{dy(t)}{(-y(t)+C)} = dt[/tex]

[tex]\int \frac{dy(t)}{(-y(t)+C)} = \int dt[/tex]

[tex]-\ln|y(t)+C| = t[/tex]

[tex]-\ln|y(t)|\ln|C|=t[/tex]

[tex]\ln\left|\frac{1}{y(t)}\right|C = t[/tex]

[tex]\frac{1}{y(t)} = e^{tC'} [/tex]

[tex]y(t) = e^{-tC'}[/tex]

[tex]y(t) = C'e^{-t}[/tex]

Therefore [tex]x(t) = ty(t) = C'te^{-t}[/tex] is another linearly independent solution.

But when I use Maple (to check that this solution works), it says that [tex]y(t) = C'e^{-t}[/tex] is not a solution. Instead is solves the ODE and finds that [tex]x(t) = t[/tex] and [tex]x(t) = \ln|t|[/tex] are the solutions. So I must have made a mistake somewhere??

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# Homework Help: Differential Equation

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