How to solve $y'+6y(t)+9\int_0^t y(\tau)d\tau=1$, $\,y(0)=0$

Question

I want to solve this equation. It reminds me something about Laplace transform.

I am sure that I must use it order to solve this:

$$y'+6y(t)+9\int_0^t y(\tau)d\tau=1,\qquad y(0)=0$$

Answer 1

Hint :

Differentiate the both side :-)

Answer 2

Let $$z(t)=\int_0^t y(\tau)d\tau$$ then the ODE becomes $$z''+6z'+9z=1,\quad z'(0)=0$$ Notice that $z_p=\frac{1}{9}$ is a particular solution and for the homogenous solution the characteristic polynomial is $$r^2+6r+9=(r+3)^2$$ and $r=-3$ is the unique solution so $$z_h(t)=(at+b)e^{-3t}$$ is the solution for the homogenous equation, and finally the general solution is $$z(t)=z_h(t)+z_p$$ The coefficients $a$ and $b$ are determined by $z(0)=0$ and $z'(0)=0$ and differentiate $z$ to find $y$.

Answer 3

If you want to use the Laplace transform, it becomes really easy:

Your equation in function of s is:

$$sY(s)+6Y(s)+\frac{9}{s}Y(s)=\frac{1}{s}$$

I used:
The derivation property: $\mathscr{L}\{y(t)'\}=sY(s)+y(0)$
The integral property: $\mathscr{L}\{\int_0^t{y(t)}\}=\frac{Y(s)}{s}$
The transformation: $\mathscr{L}\{1\}=\frac{1}{s}$

So, you obtain: $$Y(s)=\frac{1}{s^2+6s+9}=\frac{1}{(s+3)^2}$$

Now you can immediately counter transform this function to obtain:

$$y(t)=te^{-3t}$$

Using the transformation: $\mathscr{L^{-1}}\{\frac{1}{(s+a)^2}\}=te^{-at}$
or applying the frequency delay property: $\mathscr{L}\{y(t)e^{-at}\}=Y(s+a)$