Question

Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =

Answer 1

For the ODE

$ty'+2y=\sin t$

multiply both sides by t so that the left side can be condensed into the derivative of a product:

$t^2y'+2ty=t\sin t$

$\implies(t^2y)'=t\sin t$

Integrate both sides with respect to t :

$t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C$

Divide both sides by $t^2$ to solve for y :

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}$

Now use the initial condition to solve for C :

$y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}$

$\implies9=\dfrac4{\pi^2}(1+C)$

$\implies C=\dfrac{9\pi^2}4-1$

So the particular solution to the IVP is

$y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}$

or

$y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}$