Question

Solve the system of differential equations dy/dx =xz+1, dz/dx =-xy for x=0.3 using 4th order R-K method with y(0)=0, z(0)=1.

Answer 1

Solve the system of differential equations $\frac{dy}{dx} = xz + 1$ , $\frac{dz}{dx} = -xy$ for $x = 0.3$ using 4th order R-K method with $y(0) = 0$ , $z(0) = 1$ .

Solution:

Suppose we have next problem

$$\frac {d y}{d x} = f (z, x), \quad y (x _ {0}) = y _ {0}.$$

Then steps of 4th order R-K method are next

$$y _ {n + 1} = y _ {n} + \frac {1}{6} h \left(k _ {1} + k _ {2} + k _ {3} + k _ {4}\right),$$ $$x _ {n + 1} = x _ {n} + h,$$

where

$$k _ {1} = f (z _ {n}, x _ {n}),$$ $$k _ {2} = f \left(z _ {n} + \frac {1}{2} h, x _ {n} + \frac {h}{2} k _ {1}\right),$$ $$k _ {3} = f \left(z _ {n} + \frac {1}{2} h, x _ {n} + \frac {h}{2} k _ {2}\right),$$ $$k _ {4} = f \left(z _ {n} + h, x _ {n} + h k _ {3}\right).$$

In our case $x_0 = 0, y_0 = 0, z_0 = 1, h = 0.3$ . Then we have

$$y _ {1} = y _ {0} + \frac {1}{6} h \left(k _ {1} ^ {(y)} + k _ {2} ^ {(y)} + k _ {3} ^ {(y)} + k _ {4} ^ {(y)}\right);$$ $$z _ {1} = z _ {0} + \frac {1}{6} h \left(k _ {1} ^ {(z)} + k _ {2} ^ {(z)} + k _ {3} ^ {(z)} + k _ {4} ^ {(z)}\right);$$ $$k _ {1} ^ {(y)} = f ^ {(y)} \left(z _ {0}, x _ {0}\right) = x _ {0} z _ {0} + 1 = 1;$$ $$k _ {1} ^ {(z)} = f ^ {(z)} \left(y _ {0}, x _ {0}\right) = - x _ {0} y _ {0} = 0;$$ $$k _ {2} ^ {(y)} = f ^ {(y)} \left(z _ {0} + \frac {1}{2} h, x _ {0} + \frac {h}{2} k _ {1} ^ {(y)}\right) = \left(x _ {0} + \frac {h}{2} k _ {1} ^ {(y)}\right) \left(z _ {0} + \frac {1}{2} h\right) + 1 =$$ $$= \left(0 + \frac {0 . 3}{2}\right) \left(1 + \frac {0 . 3}{2}\right) + 1 = 1. 1 7 2 5;$$ $$\begin{array}{l} k_{2}^{(z)} = f^{(z)}\left(y_{0} + \frac{1}{2} h, x_{0} + \frac{h}{2} k_{1}^{(z)}\right) = -\left(x_{0} + \frac{h}{2} k_{1}^{(z)}\right)\left(y_{0} + \frac{1}{2} h\right) = \\ = -(0 + 0)\left(0 + \frac{0.3}{2}\right) = 0; \\ \end{array}$$ $$\begin{array}{l} k_{3}^{(y)} = f^{(y)}\left(z_{0} + \frac{1}{2} h, x_{0} + \frac{h}{2} k_{2}^{(y)}\right) = \left(x_{0} + \frac{h}{2} k_{2}^{(y)}\right)\left(z_{0} + \frac{1}{2} h\right) + 1 = \\ = \left(0 + \frac{0.3}{2} \cdot 1.1725\right)\left(1 + \frac{0.3}{2}\right) + 1 = 1.20225625; \\ \end{array}$$ $$\begin{array}{l} k_{3}^{(z)} = f^{(z)}\left(y_{0} + \frac{1}{2} h, x_{0} + \frac{h}{2} k_{2}^{(z)}\right) = -\left(x_{0} + \frac{h}{2} k_{2}^{(z)}\right)\left(y_{0} + \frac{1}{2} h\right) = \\ = -(0 + 0)\left(0 + \frac{0.3}{2}\right) = 0; \\ \end{array}$$ $$\begin{array}{l} k_{4}^{(y)} = f^{(y)}\left(z_{0} + h, x_{0} + h k_{3}^{(y)}\right) = \left(z_{0} + h\right)\left(x_{0} + h k_{3}^{(y)}\right) + 1 = \\ = (1 + 0.3)(0 + 0.3 \cdot 1.20225625) + 1 = 1.468879938; \\ \end{array}$$ $$\begin{array}{l} k_{4}^{(z)} = f^{(z)}\left(y_{0} + h, x_{0} + h k_{3}^{(z)}\right) = -(y_{0} + h)\left(x_{0} + h k_{3}^{(z)}\right) = \\ = -(0 + 0.3)(0 + 0.3 \cdot 0) = 0. \\ \end{array}$$

Thus we have

$$\begin{array}{l} y_{1} = y_{0} + \frac{1}{6} h\left(k_{1}^{(y)} + k_{2}^{(y)} + k_{3}^{(y)} + k_{4}^{(y)}\right) = \\ = 0 + \frac{0.3}{6} (1 + 1.1725 + 1.20225625 + 1.468879938) = 0.242181809; \\ \end{array}$$ $$\begin{array}{l} z_{1} = z_{0} + \frac{1}{6} h\left(k_{1}^{(z)} + k_{2}^{(z)} + k_{3}^{(z)} + k_{4}^{(z)}\right) = 1 + \frac{0.3}{6} \times \\ \times (0 + 0 + 0 + 0) = 1. \\ \end{array}$$

Answer:

$$\begin{array}{l} y|_{x=0.3} = 0.242181809 \\ z|_{x=0.3} = 1 \\ \end{array}$$