FIRST

\(y' = -y + ty^{1/2}, 2 \leq t \leq 3, y(2) = 2,\) with \(h = 0.25\)
\(\dfrac{d}{dy}f(t,y) = \dfrac{d}{dy} (-y + ty^{1/2}) = \dfrac{t}{2 \cdot \sqrt{y}} - 1 \\ \implies \vert \dfrac{d}{dy}f(t,y) \vert = \vert \dfrac{t}{2\cdot \sqrt{y}} - 1 \vert, \\\)
Now, since \(t \in [2,3]\), we know that this is maximized at \(t = 3 \\\) \(\implies Max_{f'} = \vert \dfrac{t}{2\cdot \sqrt{y}} - 1 \vert \\\) However, since \(y \in [-\infty, \infty]\), at \(y=0\) we get, \
\(\vert \dfrac{t}{2\cdot \sqrt{y}} - 1 \vert = \vert \dfrac{t}{2\cdot \sqrt{0}} - 1 \vert = \vert \dfrac{t}{0} - 1 \vert = \infty \\\) Thus, this problem is ill posed and doesn’t satisfy lipschitz condition.

1. Asynchronous: \

2. Asynchronous: \

3. Asynchronous: \

4. Asynchronous: \

5. Asynchronous: \

6. Asynchronous: \

7. Asynchronous: \

SECOND

1. Asynchronous: \

2. Asynchronous: \

3. Asynchronous: \

4. Asynchronous: \

5. Asynchronous: \

6. Asynchronous: \

7. Asynchronous: \

8. Asynchronous: \

THIRD

1. Asynchronous: \

2. Asynchronous: \

3. Asynchronous: \

4. Asynchronous: \

5. Asynchronous: \

6. Asynchronous: \

7. Asynchronous: \

8. Asynchronous: \

FOURTH

1. Asynchronous: \

2. Asynchronous: \

3. Asynchronous: \

4. Asynchronous: \

5. Asynchronous: \

6. Asynchronous: \

7. Asynchronous: \

8. Asynchronous: \