Let's solve **problems 3 and 4** step by step as per your requested structure. --- ## **Problem 3:** \[ y' = (y - x^2), \quad y(0) = 0, \quad h = 0.1 \] ### **Step 1: Euler's Method Calculation (All Steps)** Euler's method formula: \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \] where \( f(x, y) = (y - x^2) \). #### Initial values: - \( x_0 = 0 \) - \( y_0 = 0 \) - \( h = 0.1 \) Perform 10 steps: | Step (n) | \( x_n \) | \( y_n \) | \( f(x_n, y_n) = y_n - x_n^2 \) | \( y_{n+1} = y_n + h \cdot f(x_n, y_n) \) | |----------|-----------|-----------|----------------------------------|------------------------------------------| | 0 | 0.0 | 0.0000 | 0.0000 | 0.0000 | | 1 | 0.1 | 0.0000 | 0.0000 - (0.1)^2 = -0.0100 | 0.0000 + 0.1 \*(-0.0100) = -0.0010 | | 2 | 0.2 | -0.0010 | -0.0010 - (0.2)^2 = -0.0410 | -0.0010 + 0.1\*(-0.0410) = -0.0051 | | 3 | 0.3 | -0.0051 | -0.0051 - (0.3)^2 = -0.0951 | -0.0051 + 0.1\*(-0.0951) = -0.0146 | | 4 | 0.4 | -0.0146 | -0.0146 - (0.4)^2 = -0.1746 | -0.0146 + 0.1\*(-0.1746) = -0.0321 | | 5 | 0.5 | -0.0321 | -0.0321 - (0.5)^2 = -0.2821 | -0.0321 + 0.1\*(-0.2821) = -0.0603 | | 6 | 0.6 | -0.0603 | -0.0603 - (0.6)^2 = -0.4203 | -0.0603 + 0.1\*(-0.4203) = -0.1024 | | 7 | 0.7 | -0.1024 | -0.1024 - (0.7)^2 = -0.5924 | -0.1024 + 0.1\*(-0.5924) = -0.1617 | | 8 | 0.8 | -0.1617 | -0.1617 - (0.8)^2 = -0.8017 | -0.1617 + 0.1\*(-0.8017) = -0.2419 | | 9 | 0.9 | -0.2419 | -0.2419 - (0.9)^2 = -1.0519 | -0.2419 + 0.1\*(-1.0519) = -0.3471 | | 10 | 1.0 | -0.3471 | -0.3471 - (1.0)^2 = -1.3471 | -0.3471 + 0.1\*(-1.3471) = -0.4818 | **After 10 steps (at \(x = 1.0\))**, Euler's approximation: **\(y_{10} \approx -0.4818\)** --- ### **Step 2: Explanation of Steps** - **Initialization**: The method starts at \(x_0 = 0\) with \(y_0 = 0\). - **Iteration**: Each step computes the function value \(f(x_n, y_n)\) using the formula \(y_n - x_n^2\), then updates \(y_{n+1}\) using the Euler step. - **Update**: The value of \(y\) is decreased each step since \(x^2\) increases and dominates the expression. - **Step Size**: The increment in \(x\) is constant at 0.1. - **Process**: The process is repeated 10 times, providing the approximate solution at \(x = 1.0\). --- ### **Step 3: Conclusion** - **Euler’s Method Result**: After 10 steps, the Euler’s method approximation at \(x = 1.0\) is \(y(1.0) \approx -0.4818\). - **Trend**: The solution decreases as \(x\) increases due to the dominance of the negative \(x^2\) term. --- ## **Problem 4:** \[ y' = (y + x^2)^2, \quad y(0) = 0, \quad h = 0.1 \] ### **Step 1: Euler's Method Calculation (All Steps)** Euler's method formula: \[ y_{n+1} = y_n + h \cdot f(x_n, y_n) \] where \( f(x, y) = (y + x^2)^2 \). #### Initial values: - \( x_0 = 0 \) - \( y_0 = 0 \) - \( h = 0.1 \) Perform 10 steps: | Step (n) | \( x_n \) | \( y_n \) | \( f(x_n, y_n) = (y_n + x_n^2)^2 \) | \( y_{n+1} = y_n + h \cdot f(x_n, y_n) \) | |----------|-----------|--------------|--------------------------------------|------------------------------------------| | 0 | 0.0 | 0.000000 | (0 + 0)^2 = 0.000000 | 0.000000 | | 1 | 0.1 | 0.000000 | (0 + 0.01)^2 = 0.0001 | 0.000000 + 0.1\*0.0001 = 0.000010 | | 2 | 0.2 | 0.000010 | (0.000010 + 0.04)^2 ≈ 0.001601 | 0.000010 + 0.1\*0.001601 = 0.000170 | | 3 | 0.3 | 0.000170 | (0.000170 + 0.09)^2 ≈ 0.008117 | 0.000170 + 0.1\*0.008117 = 0.001982 | | 4 | 0.4 | 0.001982 | (0.001982 + 0.16)^2 ≈ 0.026521 | 0.001982 + 0.1\*0.026521 = 0.004634 | | 5 | 0.5 | 0.004634 | (0.004634 + 0.25)^2 ≈ 0.064808 | 0.004634 + 0.1\*0.064808 = 0.011115 | | 6 | 0.6 | 0.011115 | (0.011115 + 0.36)^2 ≈ 0.137847 | 0.011115 + 0.1\*0.137847 = 0.024900 | | 7 | 0.7 | 0.024900 | (0.024900 + 0.49)^2 ≈ 0.265350 | 0.024900 + 0.1\*0.265350 = 0.051435 | | 8 | 0.8 | 0.051435 | (0.051435 + 0.64)^2 ≈ 0.480534 | 0.051435 + 0.1\*0.480534 = 0.099488 | | 9 | 0.9 | 0.099488 | (0.099488 + 0.81)^2 ≈ 0.827070 | 0.099488 + 0.1\*0.827070 = 0.182195 | | 10 | 1.0 | 0.182195 | (0.182195 + 1.00)^2 ≈ 1.397763 | 0.182195 + 0.1\*1.397763 = 0.321971 | **After 10 steps (at \(x = 1.0\))**, Euler's approximation: **\(y_{10} \approx 0.321971\)** --- ### **Step 2: Explanation of Steps** - **Initialization**: The initial values are set as per the problem. - **Iteration**: At each step, the function \((y + x^2)^2\) is computed and used to update \(y\). - **Growth**: The solution grows faster with each step due to the squared term, which amplifies the value as both \(x\) and \(y\) increase. - **Step Size**: \(h\) remains fixed at 0.1. - **Nonlinearity**: The nonlinear nature causes rapid growth of the solution after a few steps. --- ### **Step 3: Conclusion** - **Euler’s Method Result**: After 10 steps, the Euler’s method approximation at \(x = 1.0\) is \(y(1.0) \approx 0.321971\). - **Trend**: The solution grows increasingly rapidly due to the positive feedback from the squared term. --- ## **Final Summary Statement** - For problem 3 (\(y' = y - x^2\)), the Euler method yields \(y(1.0) \approx -0.4818\) after 10 steps, showing a decreasing trend. - For problem 4 (\(y' = (y + x^2)^2\)), the Euler method yields \(y(1.0) \approx 0.321971\), with solution values increasing rapidly due to the nonlinear, squared term. **These results illustrate the application of Euler's method to both linear and nonlinear first-order ODEs, with step-by-step computational details provided for each scenario.**

Certainly! Below are the complete solutions for problems 3 and 4, presented without tables, following the requested structured format. --- ## **Problem 3:** **Differential Equation:** \[ y' = y - x^2 \] **Initial Condition:** \[ y(0) = 0 \] **Step Size:** \[ h = 0.1 \] **Number of steps:** 10 --- ### **1. Complete Solution with Calculations** **Initial values:** \[ x_0 = 0, \quad y_0 = 0 \] At each step, apply Euler's method: \[ y_{n+1} = y_n + h \times f(x_n, y_n) \] where \( f(x, y) = y - x^2 \). **Step 1:** Calculate \( f(0, 0) = 0 - 0^2 = 0 \). Update \( y \): \[ y_1 = 0 + 0.1 \times 0 = 0 \] Next \( x \): \[ x_1 = 0 + 0.1 = 0.1 \] **Step 2:** Calculate \( f(0.1, 0) = 0 - (0.1)^2 = -0.01 \). Update \( y \): \[ y_2 = 0 + 0.1 \times (-0.01) = -0.001 \] Next \( x \): \[ x_2 = 0.1 + 0.1 = 0.2 \] **Step 3:** Calculate \( f(0.2, -0.001) = -0.001 - (0.2)^2 = -0.001 - 0.04 = -0.041 \). Update \( y \): \[ y_3 = -0.001 + 0.1 \times (-0.041) = -0.001 - 0.0041 = -0.0051 \] Next \( x \): \[ x_3 = 0.2 + 0.1 = 0.3 \] **Step 4:** Calculate \( f(0.3, -0.0051) = -0.0051 - (0.3)^2 = -0.0051 - 0.09 = -0.0951 \). Update \( y \): \[ y_4 = -0.0051 + 0.1 \times (-0.0951) = -0.0051 - 0.00951 = -0.01461 \] Next \( x \): \[ x_4 = 0.3 + 0.1 = 0.4 \] **Step 5:** Calculate \( f(0.4, -0.01461) = -0.01461 - (0.4)^2 = -0.01461 - 0.16 = -0.17461 \). Update \( y \): \[ y_5 = -0.01461 + 0.1 \times (-0.17461) = -0.01461 - 0.017461 = -0.032071 \] Next \( x \): \[ x_5 = 0.4 + 0.1 = 0.5 \] **Step 6:** Calculate \( f(0.5, -0.032071) = -0.032071 - (0.5)^2 = -0.032071 - 0.25 = -0.282071 \). Update \( y \): \[ y_6 = -0.032071 + 0.1 \times (-0.282071) = -0.032071 - 0.0282071 = -0.0602781 \] Next \( x \): \[ x_6 = 0.5 + 0.1 = 0.6 \] **Step 7:** Calculate \( f(0.6, -0.0602781) = -0.0602781 - (0.6)^2 = -0.0602781 - 0.36 = -0.4202781 \). Update \( y \): \[ y_7 = -0.0602781 + 0.1 \times (-0.4202781) = -0.0602781 - 0.04202781 = -0.1023059 \] Next \( x \): \[ x_7 = 0.6 + 0.1 = 0.7 \] **Step 8:** Calculate \( f(0.7, -0.1023059) = -0.1023059 - (0.7)^2 = -0.1023059 - 0.49 = -0.5923059 \). Update \( y \): \[ y_8 = -0.1023059 + 0.1 \times (-0.5923059) = -0.1023059 - 0.05923059 = -0.1615365 \] Next \( x \): \[ x_8 = 0.7 + 0.1 = 0.8 \] **Step 9:** Calculate \( f(0.8, -0.1615365) = -0.1615365 - (0.8)^2 = -0.1615365 - 0.64 = -0.8015365 \). Update \( y \): \[ y_9 = -0.1615365 + 0.1 \times (-0.8015365) = -0.1615365 - 0.08015365 = -0.2416901 \] Next \( x \): \[ x_9 = 0.8 + 0.1 = 0.9 \] **Step 10:** Calculate \( f(0.9, -0.2416901) = -0.2416901 - (0.9)^2 = -0.2416901 - 0.81 = -1.0516901 \). Update \( y \): \[ y_{10} = -0.2416901 + 0.1 \times (-1.0516901) = -0.2416901 - 0.10516901 = -0.3468591 \] Final \( x \): \[ x_{10} = 1.0 \] --- ### **2. Explanation of Parts** - The process begins with initial conditions at \(x=0\), \(y=0\). - At each step, the function \(f(x, y) = y - x^2\) is evaluated to find the slope at the current point. - The slope is then multiplied by the step size \(h=0.1\) and added to the current \(y\) to estimate the next \(y\). - The new \(x\) value is incremented by \(h\). - This iterative process captures the approximate solution profile, with each step building upon the previous estimate. - The calculations show a consistent decrease in \(y\), reflecting the negative influence of the \( - x^2 \) term as \(x\) increases. --- ### **3. Final Conclusion** - After 10 steps, the approximate value of \( y \) at \( x=1.0 \) is approximately \(-0.34686\). - The solution indicates a steady decline in \( y \) over the interval, consistent with the differential equation's behavior. - The step-by-step calculations demonstrate the application of Euler's method for solving first-order linear differential equations numerically. --- ## **Problem 4:** **Differential Equation:** \[ y' = (y + x^2)^2 \] **Initial Condition:** \[ y(0) = 0 \] **Step Size:** \[ h = 0.1 \] **Number of steps:** 10 --- ### **1. Complete Solution with Calculations** **Initial values:** \[ x_0 = 0, \quad y_0 = 0 \] At each step, apply Euler's method: \[ y_{n+1} = y_n + h \times f(x_n, y_n) \] where \( f(x, y) = (y + x^2)^2 \). **Step 1:** Calculate \( f(0, 0) = (0 + 0)^2 = 0 \). Update \( y \): \[ y_1 = 0 + 0.1 \times 0 = 0 \] Next \( x \): \[ x_1 = 0 + 0.1 = 0.1 \] **Step 2:** Calculate \( f(0.1, 0) = (0 + 0.01)^2 = 0.0001 \). Update \( y \): \[ y_2 = 0 + 0.1 \times 0.0001 = 0.00001 \] Next \( x \): \[ x_2 = 0.1 + 0.1 = 0.2 \] **Step 3:** Calculate \( f(0.2, 0.00001) = (0.00001 + 0.04)^2 \approx (0.04001)^2 \approx 0.001601 \). Update \( y \): \[ y_3 = 0.00001 + 0.1 \times 0.001601 = 0.00001 + 0.0001601 = 0.0001701 \] Next \( x \): \[ x_3 = 0.2 + 0.1 = 0.3 \] **Step 4:** Calculate \( f(0.3, 0.0001701) = (0.0001701 + 0.09)^2 \approx (0.0901701)^2 \approx 0.008129 \). Update \( y \): \[ y_4 = 0.0001701 + 0.1 \times 0.008129 = 0.0001701 + 0.0008129 = 0.000983 \] Next \( x \): \[ x_4 = 0.3 + 0.1 = 0.4 \] **Step 5:** Calculate \( f(0.4, 0.000983) = (0.000983 + 0.16)^2 \approx (0.160983)^2 \approx 0.025908 \). Update \( y \): \[ y_5 = 0.000983 + 0.1 \times 0.025908 = 0.000983 + 0.0025908 = 0.0035738 \] Next \( x \): \[ x_5 = 0.4 + 0.1 = 0.5 \] **Step 6:** Calculate \( f(0.5, 0.0035738) = (0.0035738 + 0.25)^2 \approx (0.253574)^2 \approx 0.064297 \). Update \( y \): \[ y_6 = 0.0035738 + 0.1 \times 0.064297 = 0.0035738 + 0.0064297 = 0.0100035 \] Next \( x \): \[ x_6 = 0.5 + 0.1 = 0.6 \] **Step 7:** Calculate \( f(0.6, 0.0100035) = (0.0100035 + 0.36)^2 \approx (0.3700035)^2 \approx 0.136901 \). Update \( y \): \[ y_7 = 0.0100035 + 0.1 \times 0.136901 = 0.0100035 + 0.0136901 = 0.0236936 \] Next \( x \): \[ x_7 = 0.6 + 0.1 = 0.7 \] **Step 8:** Calculate \( f(0.7, 0.0236936) = (0.0236936 + 0.49)^2 \approx (0.5136936)^2 \approx 0.264899 \). Update \( y \): \[ y_8 = 0.0236936 + 0.1 \times 0.264899 = 0.0236936 + 0.0264899 = 0.0501835 \] Next \( x \): \[ x_8 = 0.7 + 0.1 = 0.8 \] **Step 9:** Calculate \( f(0.8, 0.0501835) = (0.0501835 + 0.64)^2 \approx (0.6901835)^2 \approx 0.476418 \). Update \( y \): \[ y_9 = 0.0501835 + 0.1 \times 0.476418 = 0.0501835 + 0.0476418 = 0.0978253 \] Next \( x \): \[ x_9 = 0.8 + 0.1 = 0.9 \] **Step 10:** Calculate \( f(0.9, 0.0978253) = (0.0978253 + 0.81)^2 \approx (0.9078253)^2 \approx 0.824252 \). Update \( y \): \[

Certainly! Continuing from the previous calculations for Problem 4: --- ### **1. Complete Solution with Calculations (continued)** **Final step (Step 10):** Calculate \( f(0.9, 0.0978253) = (0.0978253 + 0.81)^2 \approx (0.9078253)^2 \approx 0.824252 \). Update \( y \): \[ y_{10} = y_9 + h \times f(0.9, y_9) = 0.0978253 + 0.1 \times 0.824252 \approx 0.0978253 + 0.0824252 = 0.1802505 \] **At \( x = 1.0 \):** \[ \boxed{ y(1.0) \approx 0.18025 } \] --- ### **2. Explanation of Parts** - The initial condition at \(x=0\) and \(y=0\) provides the starting point for the iterative Euler method. - Each step involves calculating the function \(f(x, y) = (y + x^2)^2\), which is nonlinear and positive, leading to rapid increases in \(y\) as \(x\) increases. - The update rule \( y_{n+1} = y_n + h \times f(x_n, y_n) \) uses the current \(x, y\) to estimate the next \(y\). - The calculations show the solution's growth accelerates, especially after a few steps, due to the squared term amplifying the increase. - The step size \(h=0.1\) ensures the approximation captures the nonlinear growth reasonably well over the interval. --- ### **3. Final Conclusion** - After 10 steps, the approximate value of \( y \) at \( x=1.0 \) is approximately **0.18025**. - The nonlinear term causes the solution to grow gradually at first but accelerates as \( y \) increases, illustrating the rapid, nonlinear increase typical of squared functions. - The step-by-step calculations demonstrate the application of Euler's method to a nonlinear differential equation, providing an approximate numerical solution over the interval. --- ## **Summary of Results** - **Problem 3:** Approximate value of \( y(1.0) \) is **-0.4818**. - **Problem 4:** Approximate value of \( y(1.0) \) is **0.18025**. These solutions highlight the differences in behavior for linear versus nonlinear differential equations when solved numerically using Euler's method.