# Thread: Need urgent help on laplace transform and PDE !

1. ## Need urgent help on laplace transform and PDE !

I'm stuck with this 2 questions ...

Q1)

this is what i get after rewriting for the step function: 3sint [1-u(t-pi)] + (-3sint)u(t-pi) ... im lost from then on.

Q2)

i'm really at my wits end ... need to submit shortly afterwards. thanks a lot for the help.

2. Here is 2a.

3. 2b

4. Originally Posted by reverie414
I'm stuck with this 2 questions ...

Q1)

this is what i get after rewriting for the step function: 3sint [1-u(t-pi)] + (-3sint)u(t-pi) ... im lost from then on.

Q2)

i'm really at my wits end ... need to submit shortly afterwards. thanks a lot for the help.
I learn how to do Q1 today.

5. You expressed it correctly as a Heaviside function now you find the Laplacian of both sides.

The right hand side becomes,

[-y'(0)-sy(0)+s^2F(s)] + 4F(s) = WHATEVER.

Where, "WHATEVER" is the Laplacian of: 3sin(t) [1-u(t-pi)] -3sin(t)u(t-pi)

First you can write,
3sin(t) - 6sin(t)*u(t-pi)

Use trigonometric identity: sin(t) = -sin(t-pi)

6sin(t-pi)*u(t-pi) + 3sin(t)

Now the Laplace Transform of,
L{u(t-c)*f(t-c)} = exp(-cs) * F(s)

Thus,
L{6sin(t-pi)*u(t-pi)} = 6*exp(-pi*s)*L{sin(t)}

6*exp(-pi*s)*(1/(s^2+1^2))

So now, you have found equation in terms of Laplace transforms you just solve for F(s) an find the inverse Laplace transform.

6. thanks. may i know how do you derive this:

3sin(t) - 6sin(t)*u(t-pi)

7. Originally Posted by reverie414
thanks. may i know how do you derive this:

3sin(t) - 6sin(t)*u(t-pi)
You found this,
3sin(t) [1-u(t-pi)] -3sin(t)u(t-pi)
For the expression of the non-homogenous term.

If you open paranthesis you get,
3sin(t) - 3sin(t)*u(t-pi) - 3sin(t)u(t-pi)
Now I combine them,
3sin(t)-6sin(t)*u(t-pi)

8. oic. i get what u mean. thanks.