solutions of $u''(x)+u(x)=\frac{1}{1+x^3}$ are bounded in $\mathbb{R}^+$

Question

Prove that every solution of the equation $u''(x)+u(x)=\dfrac1{1+x^3}$ is bounded in the interval $x\in[0,\infty)$.

The problem is that a solution is hard to obtain, since the homogeneous solution is $$u_h(x)=c_1\cos x+c_2\sin x,$$ but variation of the parameters gives hard integrals to compute $$c_1'(x)=\frac{\sin x}{1+x^3},\quad c_2'(x)=\frac{\cos x}{1+x^3}.$$

Answer 1

$|c_1(x)| \leq \int_0^{\infty } \frac 1 {1+x^{3}}dx$ (plus a constant of integration) so $c_1$ is bounded. Similarly $c_2$ is also bounded. Hence $c_1(x)\cos (x) +c_2 (x) \sin (x)$ is bounded.

Answer 2

Given that the impulsive response is

$$ u_h(x)=C_1\sin x+ C_2\cos x $$

being a linear DE we have

$$ u(x) = u_h(x) * \frac{1}{1+x^2} $$

but

$$ |u_h(x) * \frac{1}{1+x^2}| \le |u_h(x) * 1| = |\int u_h dt|\le |C_1|+|C_2| $$

hence $|u(x)| \le |C_1|+|C_2|$