Use variation of parameters to find a general solution to...

Use variation of parameters to find a general solution to the differential equation given that the functions $y_1$ and $y_2$ are linearly independent solutions to the corresponding homogeneous equation for $t > 0$. $ty'' - (t+1)y' + y = 12t^2$, $y_1 = e^t$, $y_2 = t + 1$ Set up the particular solution $y_p(t) = v_1(t)y_1(t) + v_2(t)y_2(t)$ to the nonhomogeneous equation by substituting in two linearly independent solutions {$y_1(t)$, $y_2(t)$} to the corresponding homogenous equation. $y_p(t) = $

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00:05 To solve this differential equation by using the variation of parameters, what we need to do is assume y equal u1 y1 plus u2 y2, where u1 u2 are function of t.

00:34 T then we solve the linear equation in u1 prime u2 prime u1 prime y1 plus u2 prime y2 equals 0 and u1 prime y1 prime plus u2 prime y2 prime multiplied by t because we have t t here equals the right hand side 12 t squared so this is e to the t u1 prime plus t plus 1 u2 prime equal 0 and the second equation is t cancel e to the t u1 prime plus here's 1 if we differentiate this t plus 1 so this u2 prime equal 12t so u1 prime we use cramer's rule denominator it is e to the t t plus 1 e to the t 1 numerator it is 0 12t in the first column and the second column of the determinant the same as the one in denominator so t plus 1 1 so this is negative 12 t t plus 1 and the denominator it is e to the t 1 minus t plus 1 so it is negative t e to the t t negative 12 t t plus 1.

03:16 Here the negative sign cancel, t cancel, so it is 12 t plus 1 over e to the t.

03:27 So it is 12 t plus 1 e to the negative t.

03:33 For u2', it is denominated the same, so it is negative t e to the t, and the numerator, it is e to the t e to the t 0 12t.

04:07 So it is 12t e to the t over negative t e to the t, so it is negative 12.

04:29 So now we compute u1 by integration.

04:33 So it is 12 and then integral t e to the negative t dt plus integral e to the negative t.

04:49 So it is 12.

04:51 We use the integration by part so so t negative e to the negative t minus, we differentiate this t, so i keep this negative e to the negative t, so negative e to the negative t dt plus e to the negative t dt...

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