Solve this integro-differential equation.

Question

I am given the integro-differential equation $$y'(x)=\frac{7}{6} -11x +\int_0^1 (x-t) y(t)dt $$ for $x \in (0,1)$ With $$y(0)=0$$ I want to solve this by using the Laplace transform.

So I have to take the Laplace transform of both sides of the equation, the integral term is like a convolution but not quite, since we have only constant limits of integration.

A convolution would look like $$\int_0^x (x-t) y(t)dt$$.

This means that taking the Laplace transform would not help, however if I integrate both sides from $0$ to $x$ (and using the boundary condition) I get $$y(x)=\frac{7}{6} x -\frac{11x^2}{2} +\int_0^x \int_0^1 (x-t)y(t) dtds$$ By using the formula to convert a multiple integral into a single one, I end up with $$y(x)=\frac{7}{6} x -\frac{11x^2}{2} +\int_0^x (x-t)^2 y(t) dtds$$ Taking the Laplace transform in both sides I get $$\hat{y} (s)=\frac{7}{6s^2} -\frac{11}{s^3} +\frac{2}{s^3} \hat{y} $$ $$\hat{y} =\frac{7s-66}{6(s^3-2))} $$

I used Wolfram to find the inverse transform of this and gave me a huge expression, which made me believe I have made a mistake (not that huge expression are mistakes in general, but I would doubt that this is correct in this case) but I can't figure out what is the mistake, can you help?

Answer 1

If you confirm the upper bound $1$ (which is pretty dubious), you can rewrite

$$y'(x)=\frac{7}{6} -11x +\int_0^1 (x-t) y(t)dt =\frac{7}{6} -11x +x\int_0^1 y(t)dt -\int_0^1 ty(t)dt $$ and the last two integrals are some unknown constants.

Answer 2

In fact, you do not need Laplace transform:

Your equation can be written:

$$y'(x)=\frac{7}{6} -11x +x\underbrace{\int_0^1 y(t)dt}_{I}-\underbrace{\int_0^1 ty(t)dt}_J\tag{1}$$

Differentiating it gives:

$$y''(x)=-11 + I,$$

i.e. a constant.

Therefore, $$y(x)=ax^2+bx\tag{2}$$

taking into account the fact that $y(0)=0.$

It remains to plug (2) into (1)...