# Runge Kutta 2 problem

I have the following problem to solve:

Estimate y(0.1) by rk2 with h=0.1.

Where rk2 it's:

I replace the second-order ODE by a system of two first-order ODEs:

$\ \ y_{1}{}'=y_{2}\\ \ \ y_{2}{}'=48.sin(10t)-12.y_{2}-100.y_{1}\\ \ \ y_{1}(0)=0\\ \ \ y_{2}(0)=0$

Then:

$\ \ y_{1(n+1)}=y_{1(n)}+\frac{1}{2}.(k_{1}+k_{2})\\ \ \ y_{2(n+1)}=y_{2(n)}+\frac{1}{2}.(q_{1}+q_{2})$

Where, for n=0:

$\ \ k_{1}=h.y_{1}{}'(0)=h.y{}'(0)=0\\$

And now, here is my question...I am stuck at solving of this constant:

$\ \ k_{2}=h.y_{1}{}'(0.1)=h.y{}'(0.1)=?\\$

Where I get the value of $y{}'(0.1)$?.I have the same confusion when calculating the value of $q_{2} (y{}''(0.1)=?)$.Any ideas where it goes wrong?.Thanks for your help, and sorry for my poor english.

In the computation of $k$ and $q$, lose the derivatives and just use the functions $f$ and $g$ that you found by deriving the first order system, i.e.,
$f(t,y_1,y_2)=y_2$ and $g(t,y_1,y_2)=48⋅sin(10⋅t)−12⋅y_2−100⋅y_1$.