First the 2nd order ODE,

$\displaystyle \frac{d^2u}{dt^2} = 5tu + \sin (d_tu)\qquad u(0)=1 \quad d_tu=0$

Now this is what I have done, however I have no way of checking so, am I right?

$\displaystyle \begin{align*} d_tu&=v\\d_tv&=5tu+\sin (v)\\w&=\left(\begin{array}{c}u\\ v\end{array}\right)\\ d_tw&=\left(\begin{array}{c}v\\ 5tu+\sin (v)\end{array}\right)\\w(0)&=\left(\begin{array}{c }u(0)\\ v(0)\end{array}\right) = \left(\begin{array}{c}1\\ 0\end{array}\right)\end{align*}$

Thanks.