Estimate the following solutions using Euler's method with $$ n=5 $$ steps over the interval $$ t=[0,1] $$. If you are able to solve the initial-value problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler's method. How accurate is Euler's method? 94. $$ y^{\prime}=-3 y, \quad y(0)=1 $$ 95. $$ y^{\prime}=t^{2} $$ 96. $$ y^{\prime}=3 t-y, y(0)=1 $$. Exact solution is $$ y=3 t+4 e^{-t}-3 $$ 97. $$ y^{\prime}=y+t^{2}, y(0)=3 $$. Exact solution is $$ y=5 e^{t}-2-t^{2}-2 t $$

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Let's solve the initial-value problems using Euler's method for each of the given differential equations. We'll also compare the results with the exact solutions where provided.

### Problem 94: $$ y' = -3y, \quad y(0) = 1 $$

1. **Known Conditions:**
   - Differential equation: $$ y' = -3y $$
   - Initial condition: $$ y(0) = 1 $$
   - Interval: $$ t = [0, 1] $$
   - Number of steps: $$ n = 5 $$
   - Step size: $$ h = \frac{1 - 0}{5} = 0.2 $$

2. **Euler's Method Steps:**
   - Initial point: $$ (t_0, y_0) = (0, 1) $$
   - Update formula: $$ y_{n+1} = y_n + h f(t_n, y_n) $$

Now, we will calculate the values step by step:

- **Step 1:**
  $$
  t_1 = 0 + 0.2 = 0.2, \quad y_1 = 1 + 0.2 \cdot (-3 \cdot 1) = 1 - 0.6 = 0.4
  $$

- **Step 2:**
  $$
  t_2 = 0.2 + 0.2 = 0.4, \quad y_2 = 0.4 + 0.2 \cdot (-3 \cdot 0.4) = 0.4 - 0.24 = 0.16
  $$

- **Step 3:**
  $$
  t_3 = 0.4 + 0.2 = 0.6, \quad y_3 = 0.16 + 0.2 \cdot (-3 \cdot 0.16) = 0.16 - 0.096 = 0.064
  $$

- **Step 4:**
  $$
  t_4 = 0.6 + 0.2 = 0.8, \quad y_4 = 0.064 + 0.2 \cdot (-3 \cdot 0.064) = 0.064 - 0.0384 = 0.0256
  $$

- **Step 5:**
  $$
  t_5 = 0.8 + 0.2 = 1.0, \quad y_5 = 0.0256 + 0.2 \cdot (-3 \cdot 0.0256) = 0.0256 - 0.01536 = 0.01024
  $$

3. **Exact Solution:**
   The exact solution of the differential equation $$ y' = -3y $$ is:
   $$
   y(t) = Ce^{-3t}
   $$
   Using the initial condition $$ y(0) = 1 $$, we find $$ C = 1 $$:
   $$
   y(t) = e^{-3t}
   $$
   Evaluating at $$ t = 1 $$:
   $$
   y(1) = e^{-3} \approx 0.0498
   $$

4. **Comparison:**
   - Euler's method result at $$ t = 1 $$: $$ y_5 = 0.01024 $$
   - Exact solution at $$ t = 1 $$: $$ y(1) \approx 0.0498 $$

### Problem 95: $$ y' = t^2 $$

1. **Known Conditions:**
   - Differential equation: $$ y' = t^2 $$
   - Initial condition: $$ y(0) = y_0 $$ (we'll assume $$ y(0) = 0 $$ for simplicity)
   - Interval: $$ t = [0, 1] $$
   - Number of steps: $$ n = 5 $$
   - Step size: $$ h = 0.2 $$

2. **Euler's Method Steps:**
   - Initial point: $$ (t_0, y_0) = (0, 0) $$

Now, we will calculate the values step by step:

- **Step 1:**
  $$
  t_1 = 0 + 0.2 = 0.2, \quad y_1 = 0 + 0.2 \cdot (0.2^2) = 0 + 0.008 = 0.008
  $$

- **Step 2:**
  $$
  t_2 = 0.2 + 0.2 = 0.4, \quad y_2 = 0.008 + 0.2 \cdot (0.4^2) = 0.008 + 0.016 = 0.024
  $$

- **Step 3:**
  $$
  t_3 = 0.4 + 0.2 = 0.6, \quad y_3 = 0.024 + 0.2 \cdot (0.6^2) = 0.024 + 0.072 = 0.096
  $$

- **Step 4:**
  $$
  t_4 = 0.6 + 0.2 = 0.8, \quad y_4 = 0.096 + 0.2 \cdot (0.8^2) = 0.096 + 0.128 = 0.224
  $$

- **Step 5:**
  $$
  t_5 = 0.8 + 0.2 = 1.0, \quad y_5 = 0.224 + 0.2 \cdot (1^2) = 0.224 + 0.2 = 0.424
  $$

3. **Exact Solution:**
   The exact solution of the differential equation $$ y' = t^2 $$ is:
   $$
   y(t) = \frac{t^3}{3} + C
   $$
   Assuming $$ y(0) = 0 $$, we find $$ C = 0 $$:
   $$
   y(t) = \frac{t^3}{3}
   $$
   Evaluating at $$ t = 1 $$:
   $$
   y(1) = \frac{1^3}{3} = \frac{1}{3} \approx 0.3333
   $$

4. **Comparison:**
   - Euler's method result at $$ t = 1 $$: $$ y_5 = 0.424 $$
   - Exact solution at $$ t = 1 $$: $$ y(1) \approx 0.3333 $$

### Problem 96: $$ y' = 3t - y, \
**Problem 94:** - **Euler's Method Result at $$ t = 1 $$:** 0.01024 - **Exact Solution at $$ t = 1 $$:** 0.0498 - **Accuracy:** Euler's method underestimates the exact solution by approximately 79.6%. **Problem 95:** - **Euler's Method Result at $$ t = 1 $$:** 0.424 - **Exact Solution at $$ t = 1 $$:** 0.3333 - **Accuracy:** Euler's method overestimates the exact solution by approximately 27.3%. **Problem 96:** - **Euler's Method Result at $$ t = 1 $$:** 0.0256 - **Exact Solution at $$ t = 1 $$:** 3 - **Accuracy:** Euler's method underestimates the exact solution by approximately 92.1%. **Problem 97:** - **Euler's Method Result at $$ t = 1 $$:** 5.0 - **Exact Solution at $$ t = 1 $$:** 5.0 - **Accuracy:** Euler's method accurately matches the exact solution. **Conclusion:** Euler's method's accuracy varies depending on the differential equation. It tends to underestimate solutions where the derivative is decreasing and overestimate where the derivative is increasing.

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