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Abdullah's question via email about Runge Kutta scheme

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
abdullah runge kutta.jpg

First we need to write the DE as a system of first order DEs.

Let $\displaystyle u = y $ and $\displaystyle v = y' $. Then

$\displaystyle \begin{align*} x^2\,y'' - 5\,x\,v + 7\,u &= 2\,x^3\ln{\left( x \right) }\\
x^2\,y'' &= 5\,x\,v - 7\,u + 2\,x^3\ln{ \left( x \right) } \\
y'' &= \frac{5\,x\,v - 7\,u + 2\,x^3\ln{\left( x \right) } }{x^2} \end{align*} $

So the system is

$\displaystyle \begin{align*} u' &= v , \quad \quad \quad \quad \quad \quad\quad\quad \quad \quad \quad u\left( 1 \right) = 5 \\ v' &= \frac{5\,x\,v - 7\,u + 2\,x^3\ln{\left( x \right) } }{x^2} , \quad\, v\left( 1 \right) = 2 \end{align*}$

I have used my CAS to apply the Runge Kutta scheme. Here $\displaystyle f\left( x,u,v \right) = v $ and $\displaystyle g\left( x, u, v \right) = \frac{5\,x\,v - 7\,u + 2\,x^3\ln{ \left( x \right) } }{x^2} $.

abdullah runge kutta 2.jpg

abdullah runge kutta 3.jpg

abdullah runge kutta 4.jpg

abdullah runge kutta 5.jpg

After having gone through two steps of $\displaystyle h = 0.1 $, we arrive at $\displaystyle x = 1.2 $. Since we let $\displaystyle u = y $, that means $\displaystyle y\left( 1.2 \right) = u_2 = 3.28752 $.