Laplace transform - solve integral

[erba]

Homework Statement

Solve the integral

$y(t) + \int_0^t (t-u)y(u) \, du = 3sin(2t)$

Homework Equations

The Attempt at a Solution

Rewrite the equation:

$y(t) = 3sin(2t) - \int_0^t (t-u)y(u) \, du$

I assume the integral to be the convolution:

$f(t) * y(t) = t * y(t)$

as

$f(t-u) = f(t) = t$

when

$u = 0$

Laplace transform both sides, and after simplification get:

$y(s) = \frac{6s^2}{(s^2 + 4)(s^2 + 1)}$

Apply partial fraction decomposition:

$y(s) = \frac{A}{s^2 + 4} + \frac{B}{s^2 + 1}$

and I find that:

$A = 8$
$B = -2$

Thus,

$y(s) = \frac{8}{s^2 + 4} - \frac{2}{s^2 + 1}$

Inverse Laplace transform then gives:

$y(t) = 4sin(2t) - 2sin(t)$

But then I want to double-check myself by substituting y(t) into the original equation, that we were supposed to solve. I.e.:

$y(t) + \int_0^b (t-u)y(u) \, du = 3sin(2t)$

$4sin(2t) - 2sin(t) + (4sin(2t) - 2sin(t)) \int_0^t (t-u) \, du = 3sin(2t)$

$4sin(2t) - 2sin(t) + \frac{t^2}{2}(4sin(2t) - 2sin(t)) \neq 3sin(2t)$

But LHS is not equal to RHS. Where is my mistake?
My intuitive feeling is that the error lies in my "partial fraction decomposition", as all the other processes are quite straight forward.

Thank you very much!

 

[vela]

Homework Statement

Solve the integral
$$y(t) + \int_0^t (t-u)y(u) \, du = 3sin(2t)$$

Homework Equations

The Attempt at a Solution

Rewrite the equation:
$$y(t) = 3sin(2t) - \int_0^t (t-u)y(u) \, du$$I assume the integral to be the convolution:
$$f(t) * y(t) = t * y(t)$$ as
$$f(t-u) = f(t) = t$$ when $u = 0$. Laplace transform both sides, and after simplification get:
$$y(s) = \frac{6s^2}{(s^2 + 4)(s^2 + 1)}$$ Apply partial fraction decomposition:
$$y(s) = \frac{A}{s^2 + 4} + \frac{B}{s^2 + 1}$$ and I find that $A = 8$ and $B = -2$. Thus,
$$y(s) = \frac{8}{s^2 + 4} - \frac{2}{s^2 + 1}.$$ Inverse Laplace transform then gives:
$$y(t) = 4sin(2t) - 2sin(t).$$ But then I want to double-check myself by substituting y(t) into the original equation, that we were supposed to solve. I.e.:
\begin{align*}
y(t) + \int_0^b (t-u)y(u) \, du = 3sin(2t) \\
4sin(2t) - 2sin(t) + (4sin(2t) - 2sin(t)) \int_0^t (t-u) \, du = 3sin(2t)
\end{align*}

Your mistake is here. You should have y(u) and not y(t), and you can't pull it out of the integral. It should be
$$4\sin 2t - 2\sin t + \int_0^t (t-u)(4\sin 2u - 2\sin u) \, du = 3sin(2t).$$

$4sin(2t) - 2sin(t) + \frac{t^2}{2}(4sin(2t) - 2sin(t)) \neq 3sin(2t)$

But LHS is not equal to RHS. Where is my mistake?
My intuitive feeling is that the error lies in my "partial fraction decomposition", as all the other processes are quite straight forward.

Thank you very much!

 

[erba]

Your mistake is here. You should have y(u) and not y(t), and you can't pull it out of the integral. It should be
$$4\sin 2t - 2\sin t + \int_0^t (t-u)(4\sin 2u - 2\sin u) \, du = 3sin(2t).$$

Thanks for the help!

I should use y(u) as I try to solve the integral, which I now realize depends on u and not t, right?
If I would have tried to solve the convolution then y(t) would have been the right choice?

Well, when I try to solve

$4\sin 2t - 2\sin t + \int_0^t (t-u)(4\sin 2u - 2\sin u) \, du = 3sin(2t)$

I get, for the integral

$2(u-t)cos(2u) - sin(2u) + 2(u-t)cos(u) - 2sin(u)$

and I can't see how that would solve the equation. I have tried different rewritings of the cos and sin terms, but still won't end up with what's asked.
I also get the t term, which I can't get rid of.

 

[vela]

Thanks for the help!

I should use y(u) as I try to solve the integral, which I now realize depends on u and not t, right?
If I would have tried to solve the convolution then y(t) would have been the right choice?

I'm not sure what you mean. The convolution is defined as
$$(f*g)(t) = \int_{-\infty}^{\infty} f(u)g(t-u)\,du.$$ The argument of y has to be either u or t-u.

Well, when I try to solve

$4\sin 2t - 2\sin t + \int_0^t (t-u)(4\sin 2u - 2\sin u) \, du = 3sin(2t)$

I get, for the integral

$2(u-t)cos(2u) - sin(2u) + 2(u-t)cos(u) - 2sin(u)$

The last term should be the opposite sign. You need to plug in the limits now.
$$[2(u-t)\cos 2u - \sin 2u + 2(u-t)\cos u + 2\sin u]\bigg|_0^t = \ ?$$ Some of the terms will vanish when u=0 and some others when u=t.

and I can't see how that would solve the equation. I have tried different rewritings of the cos and sin terms, but still won't end up with what's asked.
I also get the t term, which I can't get rid of.

 

[erba]

I'm not sure what you mean. The convolution is defined as
$$(f*g)(t) = \int_{-\infty}^{\infty} f(u)g(t-u)\,du.$$ The argument of y has to be either u or t-u.The last term should be the opposite sign. You need to plug in the limits now.
$$[2(u-t)\cos 2u - \sin 2u + 2(u-t)\cos u + 2\sin u]\bigg|_0^t = \ ?$$ Some of the terms will vanish when u=0 and some others when u=t.

Tanks for the sign error. And I realized that I was a bit unclear in my previous post.
Solving the integral gives me:
$$-sin(2t) + 2sin(t) + 4t$$
Plugging that into the equation that we were supposed to solve gives
$$y(t) + \int_{0}^{t} (t-u)y(u)\,du = 3sin(2t)$$
$$4sin(2t) - 2sin(t) + \int_{0}^{t} (t-u)y(u)\,du = 3sin(2t)$$
$$4sin(2t) - 2sin(t) + (-sin(2t) + 2sin(t) + 4t) = 3sin(2t)$$
$$3sin(2t) + 4t = 3sin(2t)$$
$$4t = 0, t = 0$$
$$4t \neq 0, t \neq 0$$
But I guess my solution is incorrect. The solution should be 0 = 0 for all t.

 

[vela]

You have another sign error which I didn't catch before. After you integrate, you should end up with
$$[2(u-t)\cos 2u - \sin 2u - 2(u-t)\cos u + 2\sin u]\bigg|_0^t = \ ?$$ This time, the terms proportional to $t$ will cancel out.

 

[erba]

You have another sign error which I didn't catch before. After you integrate, you should end up with
$$[2(u-t)\cos 2u - \sin 2u - 2(u-t)\cos u + 2\sin u]\bigg|_0^t = \ ?$$ This time, the terms proportional to $t$ will cancel out.

Thanks for the help! Guess I should re-think even the, relatively, more trivial steps more often.