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MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: Coollight on September 27, 2019, 02:07:43 PM

Title: Q1: TUT0801
Post by: Coollight on September 27, 2019, 02:07:43 PM
Find the general solution of the given equation by using variation of parameter:

\begin{align*}
\  y' + \frac{1}{t} y = 3 cos(2t), t > 0 \\
\end{align*}
Solution:
\begin{align*}
\ y = u(t) e^ {-\int\frac{1}{t}dt} = u(t)e^ {-\ln(t)} &= u(t)t^{-1} \\
\\
\ Sub \ y = u(t)t^{-1} \ into \ the \ equation \ to \ get: \\
\ (u(t)t^{-1})' + \frac{1}{t} u(t)t^{-1} &= 3 cos(2t) \\
\ -t^{-2}u(t) + t^{-1}u(t)' + t^{-2}u(t) &= 3 cos(2t) \\
\ t^{-1}u(t)' &= 3 cos(2t) \\
\ u(t)' &= 3tcos(2t) \\
\ u(t) &= \int 3t\ cos(2t)dt \\
\ Using \ Integral \ by \ parts, & \\
\ n = 3t & \ \ \ \ m = \frac{1}{2} sin(2t) \\
\ dn = 3dt & \ \ \ \ dm = cos(2t)dt \\
\ \int 3t \ cos(2t)dt & = nm - \int mdn \\
\ &= \frac{3tsin(2t)}{2} - \int \frac{3}{2} sin(2t)dt \\
\ &= \frac{3tsin(2t)}{2} + \frac{3cos(2t)}{4} + C \\
\ Then \ we \ will \ get: \\
\ u(t) &= \frac{3tsin(2t)}{2} + \frac{3cos(2t)}{4} + C \\
\ y &= u(t)t^{-1} \\
\ &= \frac{3sin(2t)}{2} + \frac{3cos(2t)}{4t} + \frac{C}{t} \ as \ the \ general \ solution. \\
\end{align*}