differential equation of the form $\frac{du}{dt} = u(t) + f(t)$

Question

is there any way to analytically solve a differential equation of the form $\frac{du(t)}{dt} = Ku(t) + f(t)$ where k is just a constant?

Answer 1

The trick is to solve $\dot{u}=K u$, yielding $u=u_0 exp(K t)$. Then you can replace $u_0$ by a function $g(t)$. You can then inject your new $u(t)$ in your differential equation, and you get a differential equation for $g(t)$. The rest, you can figure out yourself :)

Answer 2

The given equation $$ \frac{du(t)}{dt} = Ku(t) + f(t)$$

is a first order linear differential equation.

$$ \frac{du(t)}{dt} - Ku(t) = f(t)$$

The integrating factor is $$ e^{-Kt}$$

Thus we get, $$ \frac{du(t)}{dt} e^{-Kt} - Ku(t)e^{-Kt} = e^{-Kt} f(t)$$

Integrating both sides you get $$ e^{-Kt} u(t) = \int e^{-Kt} f(t) dt + C$$

Solve for $u(t)$ and you are done.