Let's solve each differential equation step by step.

1. $\frac{dy}{dt} + 2y = e^{t/2}$, with $y(0) = 1$:

This is a first-order linear differential equation. To solve, we use the integrating factor method.

1. Standard Form:
   $\frac{dy}{dt} + 2y = e^{t/2}$

   Here, $P(t) = 2$ and $Q(t) = e^{t/2}$.

2. Integrating Factor:
   The integrating factor is given by: $\mu(t) = e^{\int P(t) \, dt} = e^{\int 2 \, dt} = e^{2t}$

3. Multiply through by the integrating factor:
   Multiply the entire equation by $e^{2t}$: $e^{2t} \frac{dy}{dt} + 2e^{2t}y = e^{2t} \cdot e^{t/2}$

   Simplify: $\frac{d}{dt} \big( e^{2t} y \big) = e^{2t} \cdot e^{t/2} = e^{(5t/2)}$

4. Integrate both sides:
   $\int \frac{d}{dt} \big( e^{2t} y \big) \, dt = \int e^{5t/2} \, dt$

   The left-hand side simplifies to $e^{2t} y$. On the right: $\int e^{5t/2} \, dt = \frac{2}{5} e^{5t/2} + C$

   So: $e^{2t} y = \frac{2}{5} e^{5t/2} + C$

5. Solve for $y$:
   $y = e^{-2t} \left( \frac{2}{5} e^{5t/2} + C \right)$

   Simplify: $y = \frac{2}{5} e^{t/2} + Ce^{-2t}$

6. Apply initial condition $y(0) = 1$:
   Substituting $t = 0$ and $y(0) = 1$: $1 = \frac{2}{5} e^{0} + Ce^{0}$

   $1 = \frac{2}{5} + C \quad \Rightarrow \quad C = 1 - \frac{2}{5} = \frac{3}{5}$

7. Final Solution:
   $y = \frac{2}{5} e^{t/2} + \frac{3}{5} e^{-2t}$

2. $\frac{dy}{dt} = \frac{ty}{t^2 + 1}$, with an unspecified initial condition:

This is a separable differential equation.

1. Rewrite in separable form:
   Divide both sides by $y$: $\frac{1}{y} \, dy = \frac{t}{t^2 + 1} \, dt$

2. Integrate both sides:
   The left-hand side becomes: $\int \frac{1}{y} \, dy = \ln|y| + C$

   The right-hand side requires substitution. Let $u = t^2 + 1$, so $du = 2t \, dt$. Then: $\int \frac{t}{t^2 + 1} \, dt = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C$

   Substitute back $u = t^2 + 1$: $\int \frac{t}{t^2 + 1} \, dt = \frac{1}{2} \ln(t^2 + 1) + C$

3. Combine results:
   $\ln|y| = \frac{1}{2} \ln(t^2 + 1) + C$

4. Exponentiate to solve for $y$:
   $y = e^C \cdot (t^2 + 1)^{1/2}$

   Let $e^C = C_1$ (a constant): $y = C_1 (t^2 + 1)^{1/2}$

5. Final Solution:
   $y = C_1 \sqrt{t^2 + 1}$

   If an initial condition is provided (e.g., $y(0) = k$), we can determine $C_1$.

Next Steps

Would you like me to:

1. Solve the second equation with specific initial conditions?
2. Provide a step-by-step verification of the solutions?
3. Clarify any part of the process above?
4. Explore applications of these differential equations?
5. Discuss the theoretical methods used here?

Tip: In linear differential equations, always verify your integrating factor calculation to ensure accuracy in solving!