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I'm reviewing for an exam, and I'm puzzled as to how to find the particular solution of the following 2nd order ODE:

y" - (2/t)y' + (2/t^2)y = 2

ICs: y(1) = 0, y'(1) = 0

I know that the complimentary solution is y = c1*t + c2*t^2

However, my attempt at finding the particular solution didn't seem to work. I haven't tried finding a particular solution for this kind of 2nd order ODE before, so I'm hoping someone could show me how?

EDIT: Wolfram Alpha shows the particular solution to be in the form t^2log(t). I'm not sure how to arrive at that, however.

Thanks!

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1 Answer 1

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We are given:

$$\tag 1 y'' - \dfrac{2}{t} y' + \dfrac{2}{t^2}2y = 2$$

Step 1

Find the homogenous solution to $(1)$, so we have:

$$\tag 2 t^2 y'' - 2 t y' + 2y = 0$$

This yields (this is a Euler-Cauchy type equation, so use $y = t^m$):

$$y_h(t) = c_1 t + c_2 t^2$$

Step 2

We are now going to make use of Variation of Parameters (VoP), so we set: $y_1 = t$ and $y_2 = t^2$ from $y_h$ and $f(t) = 2$ from $(1)$.

We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(t, t^2) = t^2$.

Using VoP, we have:

$$u_1 = \int \dfrac{-y_2 f}{W(t, t^2)}~ dt = \int -2 ~ dt = -2t$$

$$u_2 = \int \dfrac{y_1 f}{W(t, t^2)} ~dt = \int \dfrac{2}{t}~dt = 2 \ln t$$

Now, $y_p$ is given by:

$$y_p = y_1 u_1 + y_2 u_2 = t(-2t) + t^2(2 \ln t) = 2~t^2(\ln t - 1)$$

Step 3

Our solution is given by:

$$y(t) = y_h(t) + y_p(t) = c_1~t + c_2~t^2 + 2~t^2(\ln t - 1) = c_1~t + c_2~t^2 + 2~t^2~\ln t$$

Note, in the previous step, we have the terms $c_2 ~t^2 - 2t^2$ and we just absorb this into one constant as $c_2~t^2$.

Step 4

Solve for the constants using the ICs $y(1) = 0, y'(1) = 0$, arriving at:

$$c_1 = 2, c_2 = -2$$

Our final solution is:

$$y(t) = 2~t - 2~t^2 + 2~t^2\ln t = 2t(1 - t + t\ln t)$$

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  • $\begingroup$ Step 5: $\to +1$ $\endgroup$
    – amWhy
    Commented Mar 20, 2014 at 12:19

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