# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-1 => Topic started by: Victor Ivrii on January 25, 2018, 08:18:13 AM

Title: Q1-T0301
Post by: Victor Ivrii on January 25, 2018, 08:18:13 AM
Find the general solution of the given equation by variation of parameter:
\begin{equation*}
y' + \frac{1}{t}y = 3 \cos (2t),\qquad t > 0.
\end{equation*}
Title: Re: Q1-T0301
Post by: Siyuan Tao on January 25, 2018, 08:54:03 AM
It is from CH2.1 #40

According to the "Variation of Parameters", $g(t) = 3\cos(2t)$ is not everywhere zero, so we can assume the solution in the form
$$y = A(t) e^{-\int \frac{1}{t}\,dt} = A(t) e^{-\ln t} = A(t) t^{-1},$$
where $A$ is now a function of $t$.
By substituting for y in the given differential equation, $A(t)$ satistfies $A′(t) = 3t\cos(2t)$.

This implies to $A(t)$ by using"integration by parts" method.

Set  $u=3t$, $dv=\cos(2t)dt, du = 3 dt$, $v = 0.5\sin(2t)$,
change the $A(t)$ into $uv-\int vdu$ form and do the integration

Then we can get $$A(t) = \frac{3\cos(2t)}{4} + \frac{3t\sin(2t)}{2} + C$$.

So the solution is
$$y = \frac{3\cos(2t)}{4t} + \frac{3\sin(2t)}{2} + \frac{C}{t}.$$