Linear 1st order differential equation

Question

I tried to solve this equation but need help from seniors what to do next... $$\frac{dq}{dt} + q = 4\cos2t ; q(0) =1 $$ Multiplying both sides by I(t) i.e. $$I(t)= e^t$$ $$e^t\frac{dq}{dt} + e^t q =4e^t \cos2t$$ $$\frac{d}{dt}(e^tq) = 4e^t\cos2t$$ Integrating both sides $$e^t q = 4\int e^t \cos2t dt $$ Now I don't know how to integrate the right side of above equation..

Answer 1

Simple use of integration by parts:

Say, $$\begin{align} \\ & I=\int e^t \cos2t \, dt \\ & =\cos2t \int e^t \, dt - \int \left\{\frac{d}{dt}(\cos2t)\int e^t \, dt\right\}\, dt \\ & =e^t\cos2t +2 \int e^t\sin2t \, dt \\ & =e^t\cos2t +2 \left[\sin2t \int e^t \, dt - \int \left\{\frac{d}{dt}(\sin2t)\int e^t \, dt\right\}\, dt\right] \\ & =e^t\cos2t +2 \left[e^t \sin2t - 2\int e^t\cos2t \, dt\right] \\ & =e^t\cos2t + 2e^t \sin2t - 4I\end{align}$$

Therefore, we have that $$5I=e^t\cos2t + 2e^t \sin2t$$

Hope this helps you.

Answer 2

If you're familiar with the definitions of sine and cosine in terms of complex exponentials: $$ \cos \theta = \frac{1}{2} (e^{i\theta} + e^{-i\theta}), \qquad \sin \theta = -\frac{i}{2} (e^{i\theta} - e^{-i\theta}), $$ then a quick detour into the complex plane allows you to solve this without integration by parts: \begin{align} \int e^t \cos 2t \,dt &= \frac{1}{2} \int e^t \left( e^{2it} + e^{-2it} \right) dt \\ &= \frac{1}{2} \int e^{(1 + 2i) t} + e^{(1 - 2i) t} dt \\ &= \frac{1}{2} \left[ \frac{e^{(1 + 2i) t}}{1 + 2i} + \frac{e^{(1 - 2i) t}}{1 - 2i}\right] \\ &= \frac{1}{2} e^t \frac{(1 - 2i) e^{2it} + (1 + 2i) e^{-2it}}{(1 + 2i)(1 - 2i)} \\ &= \boxed{\frac{1}{5} e^t \left[ \cos 2t + 2 \sin 2t \right]} \end{align} I suspect that if you're taking a class in ODEs that you'll have to start using complex numbers soon enough, so you might as well get used to it.

Answer 3

Laplace-transforming both sides of the ODE,

$$s Q (s) - q_0 + Q (s) = \dfrac{4s}{s^2+4}$$

Since the initial condition is $q_0 = 1$, we have

$$Q (s) = \dfrac{1}{s+1} + \dfrac{4s}{(s+1)(s^2+4)} = \dfrac{1}{5}\left(\dfrac{1}{s+1}\right) + \dfrac{4}{5} \left(\dfrac{s}{s^2+4}\right) + \dfrac{8}{5} \left(\dfrac{2}{s^2+4}\right)$$

Taking the inverse Laplace transform,

$$q (t) = \frac{1}{5} \, e^{-t} + \frac{4}{5} \, \cos(2t) + \frac{8}{5} \, \sin (2t)$$