Initial value problem $u'(t)+cu(t)=f(t)$

Question

Okay so I tried to solve this initial value problem.

$$u'(t)+c*u(t)=f(t) \quad \text{for} \quad t>0, \quad u(0)=0$$

I was given the hint that I should multiply both sides with a function so that I can use the product rule.

I did that by using the e-function. $e^{\int c \, dt } = \frac{1}{c}e^{ct}$

$$\frac{1}{c}e^{ct}(u'(t)+c*u(t))=\frac{1}{c}e^{ct}f(t)$$

$$\Rightarrow \int \frac{1}{c}e^{ct}u'(t)+e^{ct}u(t) \, dt=\int \frac{1}{c}e^{ct}f(t) \, dt$$

Then i used the product rule (or better said: reversed it)...

$$\Rightarrow\frac{1}{c}e^{ct}u(t) =\int \frac{1}{c}e^{ct}f(t) \, dt$$

$$\Rightarrow u(t) = \frac{\int \frac{1}{c}e^{ct}f(t)\, dt}{\frac{1}{c}e^{ct}} = \frac{\int e^{ct}f(t)\, dt}{e^{ct}} $$

Can i simplify further? I still have to show that u(t) satisfies u(0) = 0, right? But from my perspective it computes $u(0)=\frac{\int e^{c*0}f(0)\, dt}{e^{c*0}}= \int f(0)\, dt$ or am I entirely wrong here? (I don't have so much experience)

Answer 1

I would say that you have come to the simplest form. You cannot plug $t=0$ inside the integral just to compute $u(0)$; without knowing more about $f(t)$, you can't compute the integral to the function form and thus cannot plug anything in