For the following initial value problem: $$ \left\{ \begin{array}{} \dot y=ty+\sin y \\ y(0)=10^{-2} \end{array} \right. $$
I need to evaluate $y(1)$.
I thought about using the fact that $\dot y \leq ty+1$, and solve $\dot z=tz+1$, but I couldn't solve that. I was also given a hint to use the Integral form of the equation, that is $y=10^{-2}+\int_0^t(sy+\sin y)ds$, but I can't think of a way to use that.
For $y$ that small you get $\sin y\approx y$ so that you can obtain an approximate solution from $$ \frac{y'}y=t+1\implies y(t)=y(0)\exp(\tfrac12t^2+t) $$ thus $y(1)\approx 10^{-2}e^{1.5}=0.04481689$. As the error in the equation is in the scale of $10^{-6}$, the error in the solution is of the same scale, as the integration interval has length $1$. So at least the digits $0.0448$ should be correct, the next digit is $1$ or $2$ as the 6. digit is "dirty". The numerical solution $0.0448127724253...$ is exact to at least the first 10 decimal digits, which confirms the estimate.