Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz1 => Topic started 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*}

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