# Jun's question via email about Runge Kutta Scheme...

#### Prove It

##### Well-known member
MHB Math Helper

You first have to write this DE as a system of first order equations.

Note, since $\displaystyle t$ does not appear in the original DE, that means that the system will be autonomous if kept in terms of $\displaystyle t$.

Let $\displaystyle y = u$ and $\displaystyle y' = v$, then

\displaystyle \begin{align*} y'' + 4\left( y' \right) ^2 - 7\,y &= 0.1 \\ y'' + 4\,v^2 - 7\,u &= 0.1 \\ y'' &= 7\,u - 4\,v^2 + 0.1 \end{align*}

Thus the system is

\displaystyle \begin{align*} u' &= v , \quad u\left( 0 \right) = 1 \\ v' &= 7\,u - 4\,v^2 + 0.1 , \quad v\left( 0 \right) = 0 \end{align*}

So here the system has $\displaystyle f\left( u, v \right) = v$ and $\displaystyle g\left( u, v \right) = 7\,u - 4\,v^2 + 0.1$.

I have used my CAS to work through this question.

Starting with $\displaystyle t = 0$, two steps of the scheme with stepsize $h = 0.1$ means that we are at $\displaystyle t = 0.2$, and since $\displaystyle y = u$ that means $\displaystyle y\left( 0.2 \right) = u_2 = 1.12317$.