Answer :
To apply the Runge-Kutta method of fourth order to find an approximate solution at [tex]\( x = 0.1 \)[/tex] for the differential equation [tex]\( \frac{dy}{dx} = 3x + \frac{y}{2} \)[/tex] with the initial condition [tex]\( y(0) = 1 \)[/tex], we will follow these steps:
1. Define the differential equation and initial conditions:
[tex]\[ \frac{dy}{dx} = 3x + \frac{y}{2} \][/tex]
Initial condition: [tex]\( x_0 = 0 \)[/tex], [tex]\( y_0 = 1 \)[/tex]
Step size: [tex]\( h = 0.1 \)[/tex]
Endpoint: [tex]\( x = 0.1 \)[/tex]
2. Compute the slopes ([tex]\(k\)[/tex] values) for the Runge-Kutta method:
- Calculate [tex]\( k_1 \)[/tex]:
[tex]\[ k_1 = h \cdot f(x_0, y_0) = 0.1 \cdot (3 \cdot 0 + \frac{1}{2}) = 0.1 \cdot 0.5 = 0.05 \][/tex]
- Calculate [tex]\( k_2 \)[/tex]:
[tex]\[ k_2 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}\right) = 0.1 \cdot \left(3 \cdot \frac{0.1}{2} + \frac{1 + \frac{0.05}{2}}{2}\right) \][/tex]
[tex]\[ k_2 = 0.1 \cdot \left(3 \cdot 0.05 + \frac{1.025}{2}\right) = 0.1 \cdot (0.15 + 0.5125) = 0.1 \cdot 0.6625 = 0.06625 \][/tex]
- Calculate [tex]\( k_3 \)[/tex]:
[tex]\[ k_3 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}\right) = 0.1 \cdot \left(3 \cdot \frac{0.1}{2} + \frac{1 + \frac{0.06625}{2}}{2}\right) \][/tex]
[tex]\[ k_3 = 0.1 \cdot \left(3 \cdot 0.05 + \frac{1.033125}{2}\right) = 0.1 \cdot (0.15 + 0.5165625) = 0.1 \cdot 0.6665625 = 0.06665625 \][/tex]
- Calculate [tex]\( k_4 \)[/tex]:
[tex]\[ k_4 = h \cdot f(x_0 + h, y_0 + k_3) = 0.1 \cdot \left(3 \cdot 0.1 + \frac{1 + 0.06665625}{2}\right) \][/tex]
[tex]\[ k_4 = 0.1 \cdot \left(0.3 + \frac{1.06665625}{2}\right) = 0.1 \cdot (0.3 + 0.533328125) = 0.1 \cdot 0.833328125 = 0.0833328125 \][/tex]
3. Calculate the next value of [tex]\( y \)[/tex] ([tex]\( y_{1} \)[/tex]) using the [tex]\( k \)[/tex] values:
[tex]\[ y_{1} = y_{0} + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.05 + 2 \cdot 0.06625 + 2 \cdot 0.06665625 + 0.0833328125}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.05 + 0.1325 + 0.1333125 + 0.0833328125}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.3991453125}{6} \][/tex]
[tex]\[ y_{1} = 1 + 0.06652421875 \][/tex]
Thus, the approximate value of [tex]\( y \)[/tex] at [tex]\( x = 0.1 \)[/tex] is:
[tex]\[ y(0.1) \approx 1.06652421875 \][/tex]
Therefore, using the Runge-Kutta method of fourth order, we find that the approximate solution for [tex]\( y \)[/tex] at [tex]\( x = 0.1 \)[/tex] is [tex]\( y(0.1) \approx 1.0665 \)[/tex].
1. Define the differential equation and initial conditions:
[tex]\[ \frac{dy}{dx} = 3x + \frac{y}{2} \][/tex]
Initial condition: [tex]\( x_0 = 0 \)[/tex], [tex]\( y_0 = 1 \)[/tex]
Step size: [tex]\( h = 0.1 \)[/tex]
Endpoint: [tex]\( x = 0.1 \)[/tex]
2. Compute the slopes ([tex]\(k\)[/tex] values) for the Runge-Kutta method:
- Calculate [tex]\( k_1 \)[/tex]:
[tex]\[ k_1 = h \cdot f(x_0, y_0) = 0.1 \cdot (3 \cdot 0 + \frac{1}{2}) = 0.1 \cdot 0.5 = 0.05 \][/tex]
- Calculate [tex]\( k_2 \)[/tex]:
[tex]\[ k_2 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}\right) = 0.1 \cdot \left(3 \cdot \frac{0.1}{2} + \frac{1 + \frac{0.05}{2}}{2}\right) \][/tex]
[tex]\[ k_2 = 0.1 \cdot \left(3 \cdot 0.05 + \frac{1.025}{2}\right) = 0.1 \cdot (0.15 + 0.5125) = 0.1 \cdot 0.6625 = 0.06625 \][/tex]
- Calculate [tex]\( k_3 \)[/tex]:
[tex]\[ k_3 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}\right) = 0.1 \cdot \left(3 \cdot \frac{0.1}{2} + \frac{1 + \frac{0.06625}{2}}{2}\right) \][/tex]
[tex]\[ k_3 = 0.1 \cdot \left(3 \cdot 0.05 + \frac{1.033125}{2}\right) = 0.1 \cdot (0.15 + 0.5165625) = 0.1 \cdot 0.6665625 = 0.06665625 \][/tex]
- Calculate [tex]\( k_4 \)[/tex]:
[tex]\[ k_4 = h \cdot f(x_0 + h, y_0 + k_3) = 0.1 \cdot \left(3 \cdot 0.1 + \frac{1 + 0.06665625}{2}\right) \][/tex]
[tex]\[ k_4 = 0.1 \cdot \left(0.3 + \frac{1.06665625}{2}\right) = 0.1 \cdot (0.3 + 0.533328125) = 0.1 \cdot 0.833328125 = 0.0833328125 \][/tex]
3. Calculate the next value of [tex]\( y \)[/tex] ([tex]\( y_{1} \)[/tex]) using the [tex]\( k \)[/tex] values:
[tex]\[ y_{1} = y_{0} + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.05 + 2 \cdot 0.06625 + 2 \cdot 0.06665625 + 0.0833328125}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.05 + 0.1325 + 0.1333125 + 0.0833328125}{6} \][/tex]
[tex]\[ y_{1} = 1 + \frac{0.3991453125}{6} \][/tex]
[tex]\[ y_{1} = 1 + 0.06652421875 \][/tex]
Thus, the approximate value of [tex]\( y \)[/tex] at [tex]\( x = 0.1 \)[/tex] is:
[tex]\[ y(0.1) \approx 1.06652421875 \][/tex]
Therefore, using the Runge-Kutta method of fourth order, we find that the approximate solution for [tex]\( y \)[/tex] at [tex]\( x = 0.1 \)[/tex] is [tex]\( y(0.1) \approx 1.0665 \)[/tex].
