Author Topic: Day section 2.1 #18  (Read 3231 times)

Victor Ivrii

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Day section 2.1 #18
« on: January 21, 2013, 11:01:25 AM »
Please post solution:

Find the solution of the given initial value problem:
$$
ty' + 2y = \sin (t) \quad  (t > 0),\qquad  y(\pi/2) = 1.
$$
« Last Edit: January 21, 2013, 11:05:59 AM by Victor Ivrii »

Changyu Li

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Re: Day section 2.1 #18
« Reply #1 on: January 21, 2013, 12:27:01 PM »
$$
y'+\frac{2}{t} y = \frac{\sin t}{t} \\
\mu y' + \frac{2 \mu}{t} y = \mu \frac{\sin t}{t} \\
\mu = t^2 \\
\frac{d}{dt} t^2 y = t \sin t \\
t^2 y = \int t \sin t dt \\
y = \frac{\sin t - t \cos t + C}{t^2}
$$
use initial value
$$
1 = \frac{4\left(1 - 0 + C\right)}{\pi^2}\\
C= \frac{\pi^2-4}{4}\\
y = \frac{\sin t - t \cos t + \frac{\pi^2-4}{4}}{t^2}\\
y = \frac{4\sin t - 4t \cos t + \pi^2-4}{4t^2}
$$
« Last Edit: January 21, 2013, 02:33:28 PM by Victor Ivrii »