Let's carefully solve the differential equation using the RK4 (Runge-Kutta 4th order) method with the given data. --- ### Given data: - Differential equation: \[ y' = 2x^{2/3} y^2 - 2 \] - Initial condition: \[ y(2) = 1 \] - Point of interest: \[ x = 2.5 \] - Step size: \[ h = 0.1 \] --- ### Step 1: Understand the goal We want to find \( y(2.5) \) starting from \( x=2 \), \( y=1 \), moving in steps of \( 0.1 \). Number of steps: \[ \frac{2.5 - 2}{0.1} = 5 \] So, we'll perform 5 RK4 steps to get from \( x=2 \) to \( x=2.5 \). --- ### Step 2: Define the function \( f(x, y) \) \[ f(x, y) = 2x^{2/3} y^2 - 2 \] --- ### Step 3: RK4 method overview Given current \( (x_n, y_n) \), next \( y_{n+1} \) is: \[ \begin{aligned} k_1 &= h \cdot f(x_n, y_n) \\ k_2 &= h \cdot f\left(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}\right) \\ k_3 &= h \cdot f\left(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}\right) \\ k_4 &= h \cdot f(x_n + h, y_n + k_3) \\ y_{n+1} &= y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{aligned} \] --- ### Step 4: Perform the steps --- #### **Step 1: From \( x=2.0 \), \( y=1.0 \)** Calculate \( f(2, 1) \): \[ f(2, 1) = 2 \times 2^{2/3} \times 1^2 - 2 \] Calculate \( 2^{2/3} \): \[ 2^{2/3} = \left(2^{1/3}\right)^2 \approx (1.26)^2 \approx 1.587 \] So: \[ f(2, 1) = 2 \times 1.587 \times 1 - 2 = 2 \times 1.587 - 2 = 3.174 - 2 = 1.174 \] Calculate \( k_1 \): \[ k_1 = h \times f = 0.1 \times 1.174 = 0.1174 \] Calculate \( f \) at \( x + h/2 = 2 + 0.05 = 2.05 \), \( y + k_1/2 = 1 + 0.1174/2 = 1 + 0.0587 = 1.0587 \): \[ f(2.05, 1.0587) = 2 \times (2.05)^{2/3} \times (1.0587)^2 - 2 \] Compute \( 2.05^{2/3} \): \[ 2.05^{1/3} \approx 1.266 \Rightarrow (1.266)^2 \approx 1.603 \] Calculate \( (1.0587)^2 \): \[ (1.0587)^2 \approx 1.120 \] Now: \[ f \approx 2 \times 1.603 \times 1.120 - 2 = 2 \times 1.796 - 2 = 3.592 - 2 = 1.592 \] Calculate \( k_2 \): \[ k_2 = 0.1 \times 1.592 = 0.1592 \] Next, \( f \) at \( x + h/2 = 2.05 \), \( y + k_2/2 = 1 + 0.1592/2 = 1 + 0.0796 = 1.0796 \): \[ f(2.05, 1.0796) = 2 \times 1.603 \times (1.0796)^2 - 2 \] Calculate \( (1.0796)^2 \): \[ \approx 1.166 \] So: \[ f \approx 2 \times 1.603 \times 1.166 - 2 = 2 \times 1.868 - 2 = 3.736 - 2 = 1.736 \] Calculate \( k_3 \): \[ k_3 = 0.1 \times 1.736 = 0.1736 \] Finally, \( f \) at \( x + h = 2.1 \), \( y + k_3 = 1 + 0.1736 = 1.1736 \): Calculate \( 2.1^{2/3} \): \[ 2.1^{1/3} \approx 1.273, \quad (1.273)^2 \approx 1.62 \] Calculate \( y \): \[ (1.1736)^2 \approx 1.374 \] Now evaluate \( f \): \[ f(2.1, 1.1736) = 2 \times 1.62 \times 1.374 - 2 \approx 2 \times 2.226 - 2 = 4.452 - 2 = 2.452 \] Calculate \( k_4 \): \[ k_4 = 0.1 \times 2.452 = 0.2452 \] Update \( y \): \[ y_{next} = y + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) = 1 + \frac{1}{6}(0.1174 + 2 \times 0.1592 + 2 \times 0.1736 + 0.2452) \] Sum numerator: \[ 0.1174 + 0.3184 + 0.3472 + 0.2452 = 1.0282 \] So: \[ y_{next} = 1 + \frac{1.0282}{6} \approx 1 + 0.1714 = 1.1714 \] **Result after first step:** \[ x = 2.1,\quad y \approx 1.1714 \] --- #### **Step 2: From \( x=2.1 \), \( y \approx 1.1714 \)** Repeat the process: Calculate \( f(2.1, 1.1714) \): - \( 2.1^{2/3} \approx 1.62 \) (from earlier) - \( y^2 \approx (1.1714)^2 \approx 1.372 \) \[ f \approx 2 \times 1.62 \times 1.372 - 2 \approx 2 \times 2.226 - 2 = 4.452 - 2 = 2.452 \] \( k_1 = 0.1 \times 2.452 = 0.2452 \) Next, at \( x=2.15 \), \( y=1.1714 + 0.2452/2 = 1.1714 + 0.1226 = 1.294 \): Calculate \( f \): - \( 2.15^{2/3} \): \[ 2.15^{1/3} \approx 1.278, \quad (1.278)^2 \approx 1.633 \] - \( y^2 \): \( (1.294)^2 \approx 1.674 \) \[ f \approx 2 \times 1.633 \times 1.674 - 2 \approx 2 \times 2.732 - 2 = 5.464 - 2 = 3.464 \] \( k_2 = 0.1 \times 3.464 = 0.3464 \) At \( x=2.15 \), \( y=1.1714 + 0.3464/2 = 1.1714 + 0.1732 = 1.3446 \): - \( 2.15^{2/3} \approx 1.633 \) - \( y^2 \): \( (1.3446)^2 \approx 1.808 \) \[ f \approx 2 \times 1.633 \times 1.808 - 2 \approx 2 \times 2.954 - 2 = 5.908 - 2 = 3.908 \] \( k_3 = 0.1 \times 3.908 = 0.3908 \) At \( x=2.2 \), \( y=1.1714 + 0.3908 = 1.5622 \): - \( 2.2^{2/3} \): \[ 2.2^{1/3} \approx 1.301, \quad (1.301)^2 \approx 1.692 \] - \( y^2 \): \( (1.5622)^2 \approx 2.441 \) \[ f \approx 2 \times 1.692 \times 2.441 - 2 \approx 2 \times 4.132 - 2 = 8.264 - 2 = 6.264 \] \( k_4 = 0.1 \times 6.264 = 0.6264 \) Update \( y \): \[ y_{next} = 1.1714 + \frac{1}{6}(0.2452 + 2 \times 0.3464 + 2 \times 0.3908 + 0.6264) \] Sum numerator: \[ 0.2452 + 0.6928 + 0.7816 + 0.6264 = 2.345999 \approx 2.346 \] \[ y_{next} = 1.1714 + \frac{2.346}{6} \approx 1.1714 + 0.391 = 1.5624 \] --- #### **Step 3: From \( x=2.2 \), \( y \approx 1.5624 \)** Repeat similarly: - \( 2.2^{2/3} \approx 1.692 \) - \( y^2 \): \( (1.5624)^2 \approx 2.441 \) \[ f \approx 2 \times 1.692 \times 2.441 - 2 \approx 2 \times 4.132 - 2 = 8.264 - 2 = 6.264 \] \( k_1 = 0.6264 \) At \( x=2.25 \), \( y = 1.5624 + 0.6264/2 = 1.5624 + 0.3132 = 1.8756 \): - \( 2.25^{2/3} \): \[ 2.25^{1/3} \approx 1.317, \quad (1.317)^2 \approx 1.735 \] - \( y^2 \): \( (1.8756)^2 \approx 3.518 \) \[ f \approx 2 \times 1.735 \times 3.518 - 2 \approx 2 \times 6.097 - 2 = 12.194 - 2 = 10.194 \] \( k_2 = 0.1 \times 10.194 = 1.0194 \) At \( x=2.25 \), \( y = 1.5624 + 1.0194/2 = 1.5624 + 0.5097 = 2.0721 \): - \( 2.25^{2/3} \approx 1.735 \) - \( y^2 \): \( (2.0721)^2 \approx 4.294 \) \[ f \approx 2 \times 1.735 \times 4.294 - 2 \approx 2 \times 7.455 - 2 = 14.91 - 2 = 12.91 \] \( k_3 = 0.1 \times 12.91 = 1.291 \) At \( x=2.3 \), \( y= 1.5624 + 1.291 = 2.8534 \): - \( 2.3^{2/3} \): \[ 2.3^{1/3} \approx 1.319, \quad (1.319)^2 \approx 1.743 \] - \( y^2 \): \( (2.8534)^2 \approx 8.151 \) \[ f \approx 2 \times 1.743 \times 8.151 - 2 \approx 2

Let's solve the initial value problem using the **RK4 (Runge-Kutta 4th order) method** step by step. ### **Given:** - Differential equation:  \[ y' = 2x^2 \cdot 3y^2 - 2 \] - Initial condition:  \[ y(2) = 1 \] - Step size:  \( h = 0.1 \) - Find:  \( y(2.5) \) --- ## **Step 1: Simplify the Differential Equation** \[ y' = 2x^2 \cdot 3y^2 - 2 = 6x^2 y^2 - 2 \] Let \( f(x, y) = 6x^2 y^2 - 2 \). --- ## **Step 2: Prepare for RK4 Steps** We need to go from \( x = 2 \) to \( x = 2.5 \) in steps of \( h = 0.1 \): - Points: 2.0, 2.1, 2.2, 2.3, 2.4, 2.5 - Number of steps: 5 For each step: \[ \begin{align*} k_1 & = h f(x_n, y_n) \\ k_2 & = h f(x_n + h/2, y_n + k_1/2) \\ k_3 & = h f(x_n + h/2, y_n + k_2/2) \\ k_4 & = h f(x_n + h, y_n + k_3) \\ y_{n+1} & = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \\ \end{align*} \] --- ## **Step 3: RK4 Calculations** ### **Step 1: \( x_0 = 2.0, y_0 = 1.0 \)** #### **Compute \( k_1 \):** \[ k_1 = 0.1 \cdot (6 \cdot (2.0)^2 \cdot (1.0)^2 - 2) = 0.1 \cdot (6 \cdot 4 \cdot 1 - 2) = 0.1 \cdot (24 - 2) = 0.1 \cdot 22 = 2.2 \] #### **Compute \( k_2 \):** \[ x_0 + h/2 = 2.05, \quad y_0 + k_1/2 = 1.0 + 1.1 = 2.1 \] \[ k_2 = 0.1 \cdot (6 \cdot (2.05)^2 \cdot (2.1)^2 - 2) \] First, \( (2.05)^2 = 4.2025 \), \( (2.1)^2 = 4.41 \) \[ 6 \cdot 4.2025 \cdot 4.41 = 6 \cdot 18.524 = 111.144 \] \[ k_2 = 0.1 \cdot (111.144 - 2) = 0.1 \cdot 109.144 = 10.9144 \] #### **Compute \( k_3 \):** \[ y_0 + k_2/2 = 1.0 + 10.9144/2 = 1.0 + 5.4572 = 6.4572 \] \[ k_3 = 0.1 \cdot (6 \cdot (2.05)^2 \cdot (6.4572)^2 - 2) \] First, \( (6.4572)^2 \approx 41.704 \) \[ 6 \cdot 4.2025 \cdot 41.704 = 6 \cdot 175.278 \approx 1,051.67 \] \[ k_3 = 0.1 \cdot (1,051.67 - 2) = 0.1 \cdot 1,049.67 = 104.967 \] #### **Compute \( k_4 \):** \[ x_0 + h = 2.1, \quad y_0 + k_3 = 1.0 + 104.967 = 105.967 \] \[ k_4 = 0.1 \cdot (6 \cdot (2.1)^2 \cdot (105.967)^2 - 2) \] First, \( (2.1)^2 = 4.41 \), \( (105.967)^2 \approx 11,234.9 \) \[ 6 \cdot 4.41 \cdot 11,234.9 = 6 \cdot 49,573.89 \approx 297,443.3 \] \[ k_4 = 0.1 \cdot (297,443.3 - 2) = 0.1 \cdot 297,441.3 = 29,744.13 \] #### **Update \( y_1 \):** \[ y_1 = 1.0 + \frac{1}{6}(2.2 + 2 \cdot 10.9144 + 2 \cdot 104.967 + 29,744.13) \] \[ = 1.0 + \frac{1}{6}(2.2 + 21.8288 + 209.934 + 29,744.13) \] \[ = 1.0 + \frac{1}{6}(2.2 + 21.8288 + 209.934 + 29,744.13) \] \[ = 1.0 + \frac{1}{6}(2.2 + 21.8288 + 209.934 + 29,744.13) = 1.0 + \frac{1}{6}(29,978.0928) \] \[ = 1.0 + 4,996.3488 \approx 4,997.35 \] --- ### **Analysis and Conclusion** - The **numbers increase explosively** after just one step. This is because the function \( f(x, y) = 6x^2y^2 - 2 \) grows extremely rapidly with \( y \). - **Numerical instability:** With even a modest starting value, the RK4 method gives results that explode to huge numbers in just one step. - **Interpretation:** This ODE is very "stiff" and may not be suitable for a standard RK4 with such a large step size. Reducing \( h \) might help, but the explosive nonlinearity (quadratic in \( y \)) makes it numerically challenging for explicit methods. --- ## **Summary Table of Steps** (First step only, since the values explode): | Step | \( x \) | \( y \) | \( k_1 \) | \( k_2 \) | \( k_3 \) | \( k_4 \) | Next \( y \) | |------|----------|----------|-----------|-----------|-----------|-----------|--------------| | 0 | 2.0 | 1.0 | 2.2 | 10.9144 | 104.967 | 29744.13 | 4997.35 | --- ## **Conclusion** - **RK4 for this equation and step size is not practical**: The solution blows up after the first step, so you cannot get a meaningful value for \( y(2.5) \) in this way. - **Reason**: The ODE is highly nonlinear and solutions grow extremely fast. - **Alternative**: For such rapidly growing equations, try: - A much smaller step size - A stiff solver or implicit method - Analytical methods (if possible) - Use software with adaptive step sizing (like MATLAB, Mathematica, or Python's `scipy.integrate.solve_ivp` with 'Radau' or 'BDF' methods). ### **Final Note** With the given initial condition and step size, the solution via RK4 is unstable and grows too fast to be meaningful. The solution is said to "blow up" (become infinite) very quickly due to the quadratic nonlinearity. For this ODE, a different approach or much smaller \( h \) is required. --- **If you want to try with a much smaller \( h \) or use an implicit method, let me know!**