Laplace Transform: Solving for f(x) and Finding Area Under a Curve

[acer400]

So the question is: A curve rise from the origin of the xy plane into the 1st quadrant. The area under the curve from (0,0) to (x,y) is 1/5 the area of the rectangle with these points as opposite vertices.

So I'm solving for f(x):
So far what i have is:

Area(D)=1/5 xy=integral 0 to x y(t)dt
and then

rewrite as: 1/5xy = integral 0 to x g(x-t)y(t) dt, where g(t)=1

then the next step i get stuck, because when i take the Laplace of both sides i get: 1/5L[xy]=L[y]

Thanks in advance for any help.

 

[hunt_mat]

So in LaTeX you have:
$$\int_{0}^{x}f(u)du=\frac{xf(x)}{5}$$
You know that you can differentiate to get and ODE don't you? Or do you have to use Laplace transforms?

 

[acer400]

unfortunately i must use Laplace transforms to solve. :(

 

[hunt_mat]

According to tables:
$$\mathscr{L}(xf(x))=-\mathscr{L}(f)'(s)$$ and
$$\mathscr{L}\left(\int_{0}^{x}f(t)dt\right) =\frac{\mathscr{L}(f)}{s}$$

 

[acer400]

thank you, but i still have no idea how to proceed from here.?

 

[hunt_mat]

You obtain a differential equation for the Laplace transform, solve this ODE and then invert the answer.

 

[acer400]

Can you please check if this is correct:

after applying the process of Laplace i get this, -1/5 y'-y=0. and now i just solve this ODE?

 

[hunt_mat]

No, there is an s floating around (s being the laplace transform variable) which comes from taking the laplace transform of the integral.