# Differential equation inequality help

Let $x$ be the solution to the following differential equation : $$x'(t)=\sin(t^2x)\>,\>x(0)=0.1$$ Show that

$(a)\quad \left|x(t)\right|\le0.1+t\quad\forall t>0$

$(b)\quad \left|x(t)\right|\le0.1e^\frac{t^3}{3}\quad\forall t>0$

$(a)$ I need help in verifying my solution. Using the formula : $$\text{For the IVP}:x'(t)=f(t,x)\>,\>x(t_0)=x_0\\x(t)=x_0+\int_{t_0}^{t}f(s,x(s))\,ds$$ We have \begin{align} \left|x(t)\right|&=\left|0.1+\int_{0}^{t}\sin(s^2x(s))\,ds\right|\\\\ &\le 0.1+\int_{0}^{t}\left|\sin(s^2x(s))\right|\,ds\\\\ &\le 0.1+\int_{0}^{t}1\,ds\\\\ &=0.1+t\quad\forall t>0 \end{align} Another doubt is that in the solution provided, there is this step : $$\text{Note that }\left|\sin(t^2x_1)-\sin(t^2x_2)\right|\le t^2\left|x_1-x_2\right|\quad\forall t,x_1,x_2\in\mathbb{R}$$ How is it relevant in showing the inequality?

$(b)$ \begin{align} \left|x(t)\right|&=\left|0.1+\int_{0}^{t}\sin(s^2x(s))\,ds\right|\\\\ &\le 0.1+\int_{0}^{t}\left|\sin(s^2x(s))\right|\,ds\\\\ &\le 0.1+\int_{0}^{t}\left|s^2x(s)\right|\,ds\\\\ \end{align} Using Gronwall's inequality, we have \begin{align} \left|x(t)\right|&\le0.1+\int_{0}^{t}\left|s^2x(s)\right|\,ds\\\\ &\le 0.1+\int_{0}^{t}0.1s^2e^{\left|\int_{s}^{t}z^2\,dz\right|}\,ds\\\\ &\le 0.1+\int_{0}^{t}0.1s^2e^{\int_{s}^{t}z^2dz}\,ds\\\\ &\le 0.1e^{\int_{0}^{t}z^2\,dz}\\\\ &= 0.1e^\frac{t^3}{3}\quad\forall t>0 \end{align}

Using Gronwall inequality, since $\left|x(t)\right|\le0.1+\int_{0}^{t}\left|s^2x(s)\right|\,ds$, we have
$$|x(t)|\le 0.1\exp\left(\int_0^ts^2\right)=0.1\exp\left(\frac{t^3}{3}\right)$$