Laplace Transform: $g(x)=a\sin(x)+\int_0^x \sin(x-u)g(u) du$

Question

$$g(x)=a\sin(x)+\int_0^x \sin(x-u)g(u)du$$

I need to find $g(x)$

I believe I need to use Laplace Transform with this in mind (Convolution Thm): $$(f*g)(x)= \int_0^x f(x-t)g(t)dt$$

However I don't understand how to do this problem.

Answer 1

So we have $$ g = a \cdot\mathord\sin + \mathord\sin * g $$ With $\def\L{\mathscr L}\L$ denoting the Laplace transform, we have $$ \L(g) = \L(a \cdot\mathord\sin + \mathord\sin * g) = a\L(\sin) + \L(\sin)\L(g) $$ as $\L(f*g) = \L(f)\L(g)$, hence $$ \L(g)\bigl(1-\L(sin)\bigr) = a\L(\sin) \iff \L(g) = \frac {a\L(sin)}{1-\L(\sin)} $$ As $\L(\sin)(s) = \frac 1{1+s^2}$, we have $1-\L(\sin)(s) = \frac{1+s^2-1}{1+s^2}$ $$ \L(g)(s) = \frac{a (1+s^2)}{(1+s^2)\cdot s^2} = \frac{a}{s^2} $$ Hence $g(t) = at$.