3.45. Solve the integral equation:
a) $y(\tau)=\int_{0}^{t} y(\tau) d \tau+1$;
b) $\int_{0}^{t} y(\tau) \sin (t-\tau) d \tau=1-\cos t$.
Solution.

a) We construct the image equation:

$$
\bar{y}(p)=\frac{\bar{y}(p)}{p}+\frac{1}{p} \Rightarrow \bar{y}(p)(p-1)=1 \Rightarrow \bar{y}(p)=\frac{1}{p-1}
$$

Therefore, the original function is $y(t)=e^{t}$.

b) The left side of the equation is the convolution of the function $y(t)$ and $\sin t$. Transitioning to the images, we get:

$$
\bar{y}(p) \frac{1}{p^{2}+1}=\frac{1}{p}-\frac{p}{p^{2}+1}=\frac{1}{p\left(p^{2}+1\right)} \Rightarrow \bar{y}(p)=\frac{1}{p}
$$

Therefore, $y(t)=1$.