Question

Use the Runge-Kutta method to find approximate values of the solution of the given initial value problem at the points \(x_{i}=x_{0}+i h,\) where \(x_{0}\) is the point where the initial condition is imposed and \(i=1,2\). $$ y^{\prime}=\frac{1+x}{1-y^{2}}, \quad y(2)=3 ; \quad h=0.1 $$

Short answer

Answer: For the given initial value problem, the approximate values of the solution at \(x=2.1\) and \(x=2.2\) are \((2.1, 2.9901)\) and \((2.2, 2.9804)\), respectively.

Step-by-step solution

Identify the given information

In this exercise, we are given the following information: Differential equation: \(y'=\frac{1+x}{1-y^2}\) Initial point: \((2,3)\) Step size: \(h=0.1\)

Write down the Runge-Kutta method formulas

The Runge-Kutta method (specifically, the 4th order one) involves four increment functions, usually denoted as \(k_1, k_2, k_3\) and \(k_4\). These are used to update the values of \(x\) and \(y\). The formulas for the 4th order Runge-Kutta method are: \(k_1=h\cdot f(x_n,y_n)\) \(k_2=h\cdot f(x_n+\frac{h}{2},y_n+\frac{k_1}{2})\) \(k_3=h\cdot f(x_n+\frac{h}{2},y_n+\frac{k_2}{2})\) \(k_4=h\cdot f(x_n+h,y_n+k_3)\) \(y_{n+1}=y_n+\frac{1}{6}(k_1+2k_2+2k_3+k_4)\) We will use these formulas to calculate the values of \(y_1\) and \(y_2\), which are the approximate values of the solution at \(x_1\) and \(x_2\), respectively.

Calculate k1, k2, k3, k4, and y1 for x1

We first evaluate the four increments \(k_1, k_2, k_3\) and \(k_4\) for the point \((2,3)\). \(k_1 = h \cdot f(2,3) = 0.1 \cdot \frac{1+2}{1-3^2}=-0.01\) \(k_2 = h \cdot f(2+\frac{h}{2}, 3+\frac{k_1}{2}) = 0.1 \cdot f(2.05, 2.995)$$=-0.00995025\) \(k_3 = h \cdot f(2+\frac{h}{2}, 3+\frac{k_2}{2}) = 0.1 \cdot f(2.05, 2.995025)$$=-0.0099015\) \(k_4 = h \cdot f(2+h, 3+k_3) = 0.1 \cdot f(2.1, 2.9901)=-0.009804\) Now, we can use these increments to calculate \(y_1\): \(y_1 = 3+\frac{1}{6}(-0.01+2(-0.00995025)+2(-0.0099015)+(-0.009804)) \approx 2.9901\)

Calculate k1, k2, k3, k4, and y2 for x2

Now, we evaluate the four increments \(k_1, k_2, k_3\) and \(k_4\) for the point \((2.1, 2.9901)\). \(k'_1 = h \cdot f(2.1, 2.9901) = 0.1 \cdot \frac{1+2.1}{1-{(2.9901)}^2}=-0.00980403\) \(k'_2 = h \cdot f(2.1+\frac{h}{2}, 2.9901+\frac{k'_1}{2}) = 0.1 \cdot f(2.15, 2.980198)$$=-0.00965639\) \(k'_3 = h \cdot f(2.1+\frac{h}{2}, 2.9901+\frac{k'_2}{2}) = 0.1 \cdot f(2.15, 2.980224)$$=-0.00960898\) \(k'_4 = h \cdot f(2.1+h, 2.9901+k'_3) = 0.1 \cdot f(2.2, 2.9704)$$=-0.00951326\) Now, we can use these increments to calculate \(y_2\): \(y_2 = 2.9901+\frac{1}{6}(-0.00980403+2(-0.00965639)+2(-0.00960898)+(-0.00951326)) \approx 2.9804\) Now we have the approximate values of the solution at \(x_1\) and \(x_2\): \((x_1, y_1)\approx(2.1, 2.9901)\) \((x_2, y_2)\approx(2.2, 2.9804)\)