Find function $f(x)$: $f(x)=2+3\displaystyle \int \limits_0^x t\sin tf^2(t)dt$

Question

Find function $f(x)$ continuous on $\mathbb{R}$ and satifies: $$f(x)=2+3\displaystyle \int \limits_0^x t\sin tf^2(t)dt, \forall x \in \mathbb{R}$$ I'm studying about Laplace's transform, so, I tried use it.

Set $F(s)=\mathcal{L}\left\{f(x)\right\}$ and we always have $\mathcal{L} \left\{ \displaystyle \int \limits_0^x g(\tau)d\tau \right\}=\dfrac{\mathcal{L}\left\{ g(x)\right\}}{s}$.

So, $$\begin{array}{rl} F(s)=&\dfrac{2}{s}+\dfrac{3\mathcal{L}\left\{ x\sin xf^2(x)\right\}}{s}\\ =&\dfrac{2}{s}+\dfrac{3}{s}.\dfrac{d\mathcal{L}\left\{ \sin x.f^2(x)\right\}}{ds} \end{array}$$ That's all I can do. It seems unclear. Can you help me? Thank you.

Answer 1

Hint

Such a function is at least $\mathcal C^1$. So, you can directly find $f$ s.t. $$ \begin{cases} f'(x)=3x\sin(x)f(x)^2\\ f(0)=2, \end{cases} $$ which is rather easy to solve.

Answer 2

I suggest another method. We have $$ f'(x) = 3x\sin x\,f(x)^2$$which can be written as $$ \frac{d}{dx}\left(1\over f\right) = -3x\sin x$$ By integrating this equation and using the fact that $f(0) = 2$, we can find $f(x)$.