1. ## Simple Demonstration Exercise

Given:

y'(t) + a(t).y(t) = f(t), with "a" and "f" continuos in R
a(t) >= c > 0
lim (t->oo) f(t) = 0

Demonstrate:

Any solution y(t), verifies lim (t->oo) y(t) = 0

Thank You.

2. ## Re: Simple Demonstration Exercise

Originally Posted by Pedro

Given:

y'(t) + a(t).y(t) = f(t), with "a" and "f" continuos in R
a(t) >= c > 0
lim (t->oo) f(t) = 0

Demonstrate:

Any solution y(t), verifies lim (t->oo) y(t) = 0

Thank You.
Here is a hint to get you started:

The integrating factor for this equation is

$I(t)=e^{\int_{b}^{t}a(x)dx}$

Multiplying by this and writing the left hand side as a derivative gives

$\frac{d}{dt}\left(y(t)\cdot e^{\int_{b}^{t}a(x)dx}\right)=e^{\int_{b}^{t}a(x)d x}f(t)$

Now use some of the assumptions to bound the solution to the equation.

3. ## Re: Simple Demonstration Exercise

Originally Posted by TheEmptySet

$\frac{d}{dt}\left(y(t)\cdot e^{\int_{b}^{t}a(x)dx}\right)=e^{\int_{b}^{t}a(x)d x}f(t)$

Now use some of the assumptions to bound the solution to the equation.
I've tried doing that. Maybe I'am missing something, but I'll get:

Lim [ d/dt(y(t) * e^int{a(t).dt}) ] = 0,
which gives that the integrand is constant,
C = y(t) * e^int{a(t).dt}

How can i reach that "y(t) = 0" when "t-->oo"?

4. ## Re: Simple Demonstration Exercise

Originally Posted by Pedro
I've tried doing that. Maybe I'am missing something, but I'll get:

Lim [ d/dt(y(t) * e^int{a(t).dt}) ] = 0,
which gives that the integrand is constant,
C = y(t) * e^int{a(t).dt}

How can i reach that "y(t) = 0" when "t-->oo"?
How did you get that?

If you integrate what I gave you above you should get

$y(t)\cdot e^{\int_{b}^{t}a(x)dx}=\int_{c}^{t}e^{\int_{b}^{y} a(x)dx}f(y)dy$

Solving for $y(t)$ gives

$y(t)=\frac{\int_{c}^{t}e^{\int_{b}^{y}a(x)dx}f(y)d y}{e^{\int_{b}^{t}a(x)dx}}$