To solve for all possible functions with the given derivatives, we need to integrate each derivative with respect to $t$.

a. $y' = \sin(2t)$

The general solution for $y(t)$ is obtained by integrating $\sin(2t)$:

$y(t) = \int \sin(2t) \, dt$ Using a substitution:

• Let $u = 2t$, so $du = 2 \, dt$, or $dt = \frac{du}{2}$.

Now integrate: $y(t) = \frac{1}{2} \int \sin(u) \, du = -\frac{1}{2} \cos(u) + C = -\frac{1}{2} \cos(2t) + C$ Thus, the general solution is: $y(t) = -\frac{1}{2} \cos(2t) + C$

b. $y' = \cos\left(\frac{t}{2}\right)$

Now, integrate $\cos\left(\frac{t}{2}\right)$ to find $y(t)$:

$y(t) = \int \cos\left(\frac{t}{2}\right) \, dt$ Using substitution:

• Let $u = \frac{t}{2}$, so $du = \frac{1}{2} \, dt$, or $dt = 2 \, du$.

Now integrate: $y(t) = 2 \int \cos(u) \, du = 2 \sin(u) + C = 2 \sin\left(\frac{t}{2}\right) + C$ Thus, the general solution is: $y(t) = 2 \sin\left(\frac{t}{2}\right) + C$

c. $y' = \sin(2t) + \cos\left(\frac{t}{2}\right)$

The general solution is the sum of the two integrals from parts a and b:

$y(t) = \int \left( \sin(2t) + \cos\left(\frac{t}{2}\right) \right) \, dt$ From part a, we know: $\int \sin(2t) \, dt = -\frac{1}{2} \cos(2t)$ From part b, we know: $\int \cos\left(\frac{t}{2}\right) \, dt = 2 \sin\left(\frac{t}{2}\right)$

Thus, the general solution is: $y(t) = -\frac{1}{2} \cos(2t) + 2 \sin\left(\frac{t}{2}\right) + C$

Summary of Solutions:

a. $y(t) = -\frac{1}{2} \cos(2t) + C$

b. $y(t) = 2 \sin\left(\frac{t}{2}\right) + C$

c. $y(t) = -\frac{1}{2} \cos(2t) + 2 \sin\left(\frac{t}{2}\right) + C$

Would you like any further details or explanations?
Here are some related questions you might find useful:

1. What is the importance of the constant $C$ in the general solution?
2. How does substitution work in integration problems like these?
3. Can you explain how the chain rule applies to these integrals?
4. What is the relationship between derivatives and antiderivatives?
5. How can we verify that our solutions are correct?

Tip: Always double-check your integral using differentiation to confirm the solution is correct.