Question

Solve the system of differentil 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

$$\left\{ \begin{array}{l} \frac {d y}{d x} = x z + 1 \\ \frac {d z}{d x} = - x y \end{array} \right.,$$ $$y (0) = 0, z (0) = 1, x = 0. 3$$

Solution

Solve this system with a step $h = 0.3$, $\varphi(x, y, z) = xz + 1, \psi(x, y, z) = -xy$, $x_0 = 0$, $y_0 = 0$, $z_0 = 1$

$$y _ {0. 3} = y _ {0 + 0. 3} = y _ {0} + \frac {h}{6} \left(k _ {1} + 2 k _ {2} + 2 k _ {3} + k _ {4}\right),$$ $$z _ {0. 3} = z _ {0 + 0. 3} = z _ {0} + \frac {h}{6} \left(l _ {1} + 2 l _ {2} + 2 l _ {3} + l _ {4}\right)$$ $$k _ {1} = \varphi \left(x _ {0}, y _ {0}, z _ {0}\right) = 0 \cdot 1 + 1 = 1$$ $$l _ {1} = \psi \left(x _ {0}, y _ {0}, z _ {0}\right) = 0$$ $$k _ {2} = \varphi \left(x _ {0} + h / _ {2}, y _ {0} + h k _ {1} / _ {2}, z _ {0} + h l _ {1} / _ {2}\right) = 1. 1 5$$ $$l _ {2} = \psi \left(x _ {0} + h / _ {2}, y _ {0} + h k _ {1} / _ {2}, z _ {0} + h l _ {1} / _ {2}\right) = - 0. 0 2 2 5$$ $$k _ {3} = \varphi \left(x _ {0} + h / _ {2}, y _ {0} + h k _ {2} / _ {2}, z _ {0} + h l _ {2} / _ {2}\right) = 1. 1 4 9 4 9 3 7 5$$ $$l _ {3} = \psi \left(x _ {0} + h / _ {2}, y _ {0} + h k _ {2} / _ {2}, z _ {0} + h l _ {2} / _ {2}\right) = - 0. 0 2 5 8 7 5$$ $$k _ {4} = \varphi \left(x _ {0} + h, y _ {0} + h k _ {3}, z _ {0} + h l _ {3}\right) = 1. 2 9 7 6 7 1 2 5$$ $$l _ {4} = \psi \left(x _ {0} + h, y _ {0} + h k _ {3}, z _ {0} + h l _ {3}\right) = - 0. 1 0 3 4 5 4 4 3 7 5$$ $$y _ {0. 3} = 0 + \frac {0 . 3}{6} (1 + 2 \cdot 1. 1 5 + 2 \cdot 1. 1 4 9 4 9 3 7 5 + 1. 2 9 7 6 7 1 2 5) = 0. 3 4 4 8 3 2 9 3 7 5$$ $$z _ {0. 3} = 1 + \frac {0 . 3}{6} (0 + 2 \cdot (- 0. 0 2 2 5) + 2 \cdot (- 0. 0 2 5 8 7 5) - 0. 1 0 3 4 5 4 4 3 7 5) = 0. 9 8 9 9 8 9 7 7 8 1 2 5$$

General solution:

From the second equation: $y = -\frac{1}{x}\frac{dz}{dx}$, let differentiate it changing $\frac{dy}{dx} = xz + 1$:

$$x z + 1 = \frac {1}{x ^ {2}} \frac {d z}{d x} - \frac {1}{x} \frac {d ^ {2} z}{d x ^ {2}},$$ $$\frac {1}{x} \frac {d ^ {2} z}{d x ^ {2}} - \frac {1}{x ^ {2}} \frac {d z}{d x} + x z + 1 = 0$$

Let $t = \frac{x^2}{2}$, so $x = \sqrt{2t}$:

$$\frac {\frac {d ^ {2} z}{d x ^ {2}}}{\sqrt {2 t}} + \sqrt {2 t} z - \frac {\frac {d z}{d x}}{2 t} + 1 = 0,$$

By the chain rule $\frac{dz}{dx} = \frac{dt}{dx}\frac{dz}{dt}$ and $\frac{d^2z}{dx^2} = (\frac{dt}{dx})^2\frac{d^2z}{dt^2} +\frac{d^2t}{dx^2}\frac{dz}{dt}$.

Therefore $\frac{dz}{dx} = \sqrt{2t}\frac{dz}{dt}$ and $\frac{d^2z}{dx^2} = 2t\frac{d^2z}{dt^2} +\frac{dz}{dt}$.

Substitute these values into the differential equation:

$$\frac {2 t \frac {d ^ {2} z}{d t ^ {2}} + \frac {d z}{d t}}{\sqrt {2 t}} + \sqrt {2 t} z - \frac {\frac {d z}{d t}}{\sqrt {2 t}} + 1 = 0.$$

Simplify and divide both sides by $\sqrt{2t}$ :

$$\frac {d ^ {2} z}{d t ^ {2}} + z + \frac {1}{\sqrt {2 t}} = 0.$$

The general solution is a sum of the complementary and particular solutions.

Find complementary solution from: $\frac{d^2z}{dt^2} + z = 0$ . Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$ :

$$\frac {d ^ {2}}{d t ^ {2}} \left(e ^ {\lambda t}\right) + e ^ {\lambda t} = 0.$$

Substitute $\frac{d^2}{dt^2}\left(e^{\lambda t}\right) = \lambda^2 e^{\lambda t},$ so $\lambda^2 e^{\lambda t} + e^{\lambda t} = 0,$

$e^{\lambda t}(\lambda^2 + 1) = 0$ , such as $e^{\lambda t} \neq 0$ then $\lambda^2 + 1 = 0$ . Solve it: $\lambda = \pm i$ .

So $z_{c} = c_{1}\cos (t) + c_{2}\sin (t)$ .

Determine the particular solution to $\frac{d^2z}{dt^2} + z = -\frac{1}{\sqrt{2t}}$ by variation of parameters.

List the basis solutions in $z_{c}$ : $z_{b1} = \cos (t)$ and $z_{b2} = \sin (t)$ .

Compute the Wronskian of $z_{b1}$ and $z_{b2}$ :

$$W (t) = \left| \begin{array}{c c} \cos (t) & \sin (t) \\ \frac {d \cos (t)}{d t} & \frac {d \sin (t)}{d t} \end{array} \right| = \left| \begin{array}{c c} \cos (t) & \sin (t) \\ - \sin (t) & \cos (t) \end{array} \right| = 1.$$

Let $f(t) = -\frac{1}{\sqrt{2t}}$

let $\nu_{1}(t) = -\int \frac{f(t)z_{b2}}{W(t)} dt$ and $\nu_{2}(t) = -\int \frac{f(t)z_{b1}}{W(t)} dt.$

The particular solution will be given by:

$$z _ {p} = \nu_ {1} (t) z _ {b 1} + \nu_ {2} (t) z _ {b 2}.$$ $$\nu_ {1} (t) = - \int \frac {- \sin (t)}{\sqrt {2 t}} d t = \sqrt {\pi} S (\sqrt {\frac {2 t}{\pi}}), \nu_ {2} (t) = \int \frac {- \cos (t)}{\sqrt {2 t}} d t = - \sqrt {\pi} C (\sqrt {\frac {2 t}{\pi}}).$$

So $z_{p} = \sqrt{\pi} S\left(\sqrt{\frac{2t}{\pi}}\right)\cos (t) - \sqrt{\pi} C\left(\sqrt{\frac{2t}{\pi}}\right)\sin (t) = \sqrt{\pi}\left(S\left(\sqrt{\frac{2t}{\pi}}\right)\cos (t) - C\left(\sqrt{\frac{2t}{\pi}}\right)\sin (t)\right).$

The general solution:

$$z = z _ {c} + z _ {p} = c _ {1} \cos (t) + c _ {2} \sin (t) + \sqrt {\pi} \big (S \left(\sqrt {\frac {2 t}{\pi}}\right) \cos (t) - C \left(\sqrt {\frac {2 t}{\pi}}\right) \sin (t) \big).$$

Substitute back for $t = \frac{x^2}{2}$ :

$$z = c _ {1} \cos \left(\frac {x ^ {2}}{2}\right) + c _ {2} \sin \left(\frac {x ^ {2}}{2}\right) + \sqrt {\pi} \left(S \left(\frac {x}{\sqrt {\pi}}\right) \cos \left(\frac {x ^ {2}}{2}\right) - C \left(\frac {x}{\sqrt {\pi}}\right) \sin \left(\frac {x ^ {2}}{2}\right)\right).$$ $$\frac {d z}{d x} =$$ $$- c _ {1} \times \sin \left(\frac {x ^ {2}}{2}\right) + c _ {2} \times \cos \left(\frac {x ^ {2}}{2}\right) - \sin \left(\frac {x ^ {2}}{2}\right) C ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) - \sqrt {\pi} \times \cos \left(\frac {x ^ {2}}{2}\right) C \left(\frac {x}{\sqrt {\pi}}\right) + + \cos \left(\frac {x ^ {2}}{2}\right) S ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) - \sqrt {\pi} \times \sin \left(\frac {x ^ {2}}{2}\right) S \left(\frac {x}{\sqrt {\pi}}\right).$$ $$\text{So } y = - \frac {1}{x} \frac {d z}{d x} = c _ {1} \sin \left(\frac {x ^ {2}}{2}\right) - c _ {2} \cos \left(\frac {x ^ {2}}{2}\right) + \frac {1}{x} \sin \left(\frac {x ^ {2}}{2}\right) C ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) + \sqrt {\pi} \cos \left(\frac {x ^ {2}}{2}\right) C \left(\frac {x}{\sqrt {\pi}}\right) - \frac {1}{x} \cos \left(\frac {x ^ {2}}{2}\right) S ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) + \sqrt {\pi} \sin \left(\frac {x ^ {2}}{2}\right) S \left(\frac {x}{\sqrt {\pi}}\right).$$

**Answer**

$$y _ {0. 3} = 0. 3 4 4 8 3 2 9 3 7 5, z _ {0. 3} = 0. 9 8 9 9 8 9 7 7 8 1 2 5.$$ $$z = c _ {1} \cos \left(\frac {x ^ {2}}{2}\right) + c _ {2} \sin \left(\frac {x ^ {2}}{2}\right) + \sqrt {\pi} \left(S \left(\frac {x}{\sqrt {\pi}}\right) \cos \left(\frac {x ^ {2}}{2}\right) - C \left(\frac {x}{\sqrt {\pi}}\right) \sin \left(\frac {x ^ {2}}{2}\right)\right).$$ $$y =$$ $$c _ {1} \sin \left(\frac {x ^ {2}}{2}\right) - c _ {2} \cos \left(\frac {x ^ {2}}{2}\right) + \frac {1}{x} \sin \left(\frac {x ^ {2}}{2}\right) C ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) + \sqrt {\pi} \cos \left(\frac {x ^ {2}}{2}\right) C \left(\frac {x}{\sqrt {\pi}}\right) - \frac {1}{x} \cos \left(\frac {x ^ {2}}{2}\right) S ^ {\prime} \left(\frac {x}{\sqrt {\pi}}\right) + \sqrt {\pi} \sin \left(\frac {x ^ {2}}{2}\right) S \left(\frac {x}{\sqrt {\pi}}\right).$$