Solve the initial value problem $$U'(t) + \omega^2\alpha^2U(t) = \frac{2A\omega\alpha^2}{\pi}$$ with initial condition $$U(0) = 0$$.

Our integrating factor is $$\mu(t) = e^{\omega^2\alpha^2t}$$. Using the inverse product rule we get: $$\frac{d}{dt}\left[U(t)e^{\omega^2\alpha^2t} = \frac{2A\omega\alpha^2}{\pi}$$. From here, I need a homogeneous solution and a particular solution, but I'm not entirely sure how to proceed. My immediate reaction to to just integrate, but I don't think that gives me the correct answer.

Question

Solve the initial value problem $$U'(t) + \omega^2\alpha^2U(t) = \frac{2A\omega\alpha^2}{\pi}$$ with initial condition $$U(0) = 0$$.

Our integrating factor is $$\mu(t) = e^{\omega^2\alpha^2t}$$. Using the inverse product rule we get: $$\frac{d}{dt}\left[U(t)e^{\omega^2\alpha^2t} = \frac{2A\omega\alpha^2}{\pi}$$. From here, I need a homogeneous solution and a particular solution, but I'm not entirely sure how to proceed. My immediate reaction to to just integrate, but I don't think that gives me the correct answer.

anonymous · Answer

it looks like you only need to use integrating factors

anonymous · Answer

multiply both sides by the integrating factor:

u'(t) * e^(w^2*a^2*t) + w^2 * a^2 * u(t) * e^(w^2 * a^2 * t) = 2Awa^2/pi * e^(w^2*a^2*t)

integrate both sides with respect to t:

u(t) * e^(w^2 * a^2 * t) = 2A/(pi*w) * e^(w^2*a^2*t) + C

therefore u(t) = 2A/(pi*w) + Ce^(-w^2 * a^2 * t)

anonymous · Answer

since u(0) = 0:

0 = 2A/(pi*w) + C therefore C = -2A/(pi*w)

thus u(t) = 2A/(pi*w)[1 - e^(-w^2*a^2*t)]

nocipher · Answer

gotcha. I forgot to apply the integrating factor to both sides of the equation; That's why I kept getting a weird answer. Thanks for the help.