Use the integrating factor method to find y solution of the...

Use the integrating factor method to find $y$ solution of the initial value problem $ty' = 3y + 2t^4 \sin(3t)$, $y(\frac{\pi}{3}) = 0$ $t > 0$. (a) Find an integrating factor $\mu$. If you leave an arbitrary constant, denote it as $c$. $\mu(t) = t^{-3}$ (b) Find all solutions $y$ of the differential equation above. Again denote by $c$ any arbitrary integration constant. $y(t) = $ (c) Find the only solution of the initial value problem above. $y(t) = $

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00:01 Okay, first step here, let's rearrange this and write y dashed as dy by dt, so plus 3y equals t.

00:15 That's the first step.

00:16 Now in that format if you have, let's say, normally it's dy by dx, not too important, dy by dx plus py equals q, where p and q are functions of x in this case, then the integrating factor will always be e to the integral of p dx.

00:45 That rule, learn of by heart.

00:49 Here it's dy by dt.

00:51 Same idea though.

00:53 So my integrating factor in this case will be e to the integral of 3, in this case, dt.

01:05 In other words, it will be e to the power of 3t.

01:12 That's the first step.

01:15 Then you multiply both sides of the equation by that factor.

01:20 So what we have then, take this, i'm going to have e to the 3t times dy by dt plus 3y e to the 3t equals, on the right we have t, t, e to the 3t.

01:49 Now by doing that, the left hand side is always an exact differential.

01:57 And it will be d by dt of y e to the 3t.

02:06 If i do that differential, i end up with this result here.

02:13 Equals te to the 3t.

02:20 Then i integrate both sides, so i get y e to the 3t.

02:24 Equals the integral of t e to the 3t dt.

02:31 On the right, we do it by parts...

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