Ytukyg

Given the equation
$$
y''=ccdot(1+(y')^2)^{1/2} ~~where~~ c=0.053.
$$
Putting this in system form, I get
begin{align}
y'&=z
\
y''&=c⋅(1+z^2)^{1/2}
end{align}
I am to use 4th Order Runge-Kutta method to solve this for $y(30)$,

given initial conditions $y(0)=25$ and $y'(0)=0$.

However, I am confused as to how I incorporate $y$ into my method if $y$ is not in the given equation.

Any help would be really appreciated.

You are right that you could apply the integration only to $z=y'$, $z'=csqrt{1+z^2}$, $z(0)=y'(0)$ and then obtain $y$ via cumulative summation of the function table of $z$.

However, in treating it as a system there is no difference if $y$ occurs explicitly or not, the formulation of the vector version of RK4 takes no notice of the variable dependencies of the component functions.

If $y'=f(y,z)$ and $z'=g(y,z)$, here with $f(y,z)=z$ and $g(y,z)=h(z)$, then the RK4 stages start as

k1y = dt*f(y,z)

k1z = dt*g(y,z)



k2y = dt*f(y+0.5*k1y, z+0.5*k1z)

k2z = dt*g(y+0.5*k1y, z+0.5*k1z)

etc.

By integration, $$frac{y''}{sqrt{1+y'^2}}=c$$

gives

$$text{arsinh }y'=cx+d,$$

$$y'=sinh(cx+d)$$ and

$$y=frac1ccosh(cx+d)+e$$

where

$$sinh d=0,$$

$$frac1ccosh(d)+e=25.$$

Do you really need numerical integration ?