Runge-Kutta 4th order to solve 2nd order ODE

[jamesbellcpp]

I tried to write a code to solve a 2nd order ODE but somehow it did not work as I intended.

the equation is 2y" + 5y' +3y = 0 ,y(0) = 3 ,y'(0) = -4
The final answer will be y(x) = 3-4x
therefore y(1) = -1 and y'(1) = -4
But the program output is y(1) = -0.999995 which is somehow correct
However, y'(1) is output a wrong answer.
Please help!
Thank you!

#include <iostream>
#include <cmath>
#include <math.h> 

double ddx(double x,double y,double z)
{
	double dx = (-5*z-3*y)/2;
	return dx;
}

double test2ndorder(double x0, double y0, double z0, double x, double h)
{
	int n = (int)((x - x0) / h);
	double k1, k2, k3, k4;
	double l1, l2, l3, l4;

	double y = y0;
	double z = z0;

	for (int i = 1; i <= n; i++)
	{

		k1 = h * z0;
		l1 = h * ddx(x0, y0, z0);
		k2 = h * (z0 + 0.5*l1);
		l2 = h * ddx(x0 + 0.5*h, y0 + 0.5*k1, z0 + 0.5*l1);
		k3 = h * (z0 + 0.5*l2);
		l3 = h * ddx(x0 + 0.5*h, y0 + 0.5*k2, z0 + 0.5*l2);
		k4 = h * (z0 + l3);
		l4 = h * ddx(x0 + h, y0 + k3, z0 + l3);

		y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4);
		z = z + (1.0 / 6.0)*(l1 + 2 * l2 + 2 * l3 + l4);
		x = x + h;
	}
	return z;
}

int main()
{
	double x0, y0, z0, x, y, z,h;
	x0 = 0;
	x = 1;
	y0 = 3;
	z0 = -4;
	h =0.0001;

	z = test2ndorder(x0, y0, z0, x, h);
	std::cout << z;
}

[lastchance]

It won't compile if you start your function names with a 2.

ddx doesn't correspond to your given equation (you shouldn't divide by 2).

How do you know what dy/dx is at the end? - you don't output it. (EDIT - now you've edited your code and do output it ... but not y!)

Pass n, not h, or you will be subject to rounding error.

Please include headers.

[jamesbellcpp]

Sorry, the code I post before was an unupdate version.
Please look at the new version.
The output is 1.4614 but the answer should be -4 according to the equation.

[lastchance]

ddx doesn't correspond to the differential equation that you cite. Either your differential equation should have a 2 before the second derivative, or the result of this function shouldn't end by dividing by 2.

In neither case is the solution y=3-4x. It has a solution in decaying exponentials.

Your function should return both y and z(=dy/dx). You will need to return them through reference parameters.

[jamesbellcpp]

Thank you for the answer!

Turn out, I also use x0 ,y0 and z0 to calculate x, y and z in every step.
after I change that ,it works :D