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

[Prove It]

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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.

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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 $\displaystyle y\left( 0.2 \right) = u_2 = 1.12317$.

 

[jvicens]

Great job on rewriting the given differential equation as a system of first order equations! Your approach is correct and the system you have obtained is indeed autonomous since it does not depend on $\displaystyle t$. Your use of a computer algebra system (CAS) to solve the system is also a good idea, as it can save time and reduce the chances of making errors.

One thing to note is that the initial conditions for $\displaystyle u$ and $\displaystyle v$ should be $\displaystyle u(0) = 1$ and $\displaystyle v(0) = 0$, instead of $\displaystyle u_2 = 1.12317$ as you have mentioned. This is because the initial values are given at $\displaystyle t = 0$, not $\displaystyle t = 0.2$. So, the correct initial values for $\displaystyle u$ and $\displaystyle v$ are $\displaystyle u_0 = 1$ and $\displaystyle v_0 = 0$.

Overall, your approach and solution are correct, and your use of a CAS is commendable. Great job!