# Mahesh's question via email about a Runge-Kutta scheme

#### Prove It

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

To apply this Runge-Kutta scheme, we will need to write our second order DE as a system of first order DEs.

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

\displaystyle \begin{align*} y'' + 4\,v - 7\,u^2 &= 0.2 \\ y'' &= 0.2 - 4\,v + 7\,u^2 \end{align*}

So our system of first order DEs is:

\displaystyle \begin{align*} u' &= v , \quad \quad \quad \quad \quad \quad \quad \, u\left( 0 \right) = 3 \\ v' &= 0.2 - 4\,v + 7\,u^2 , \quad v\left( 0 \right) = 0 \end{align*}

Now we can apply the Runge-Kutta scheme. Note that $\displaystyle f\left( u,v \right) = v$ and $\displaystyle g\left( u,v \right) = 0.2 - 4\,v + 7\,u^2$, and the step size is $\displaystyle h = 0.05$.

I have used my CAS to solve this.

Moving along two steps of size $\displaystyle h = 0.05$ gets us to $\displaystyle t = 0.1$, and since $\displaystyle u = y$, that means we have $\displaystyle y\left( 0.2 \right) = u_2 = 3.28623$.