91Ó°ÊÓ

First, rewrite the given differential equation: $$ y^{\prime}-y=1+3 \sin t $$ This is a first-order linear ordinary differential equation in the form of: $$ y^{\prime} + P(t)y = Q(t) $$ Here, we have \(P(t)=-1\) and \(Q(t)=1+3\sin t\). Now, we will find the integrating factor (IF) which is given by \(e^{\int P(t) \, dt}\). Calculate the integrating factor: $$ IF = e^{\int -1 dt} = e^{-t} $$ Now, multiply both sides of the differential equation with the integrating factor: $$ e^{-t}y^{\prime} - e^{-t}y = e^{-t}(1+3\sin t) $$ Now, observe that the left-hand side of the equation is the derivative of the product \(y(t)e^{-t}\): $$ \frac{d}{dt}(y(t)e^{-t}) = e^{-t}(1+3\sin t) $$ Now, we integrate both sides with respect to t: $$ \int \frac{d}{dt}(y(t)e^{-t}) dt = \int e^{-t}(1+3\sin t) dt $$

Using integration by parts (twice) on the right-hand side of the equation and remembering that the integral of the derivative of a function is the function itself, we have: $$ y(t)e^{-t} = -e^{-t} - 3e^{-t}\cos t + C $$ To find the general solution, multiply both sides by \(e^t\) to clear out the exponential term: $$ y(t) = -1 - 3\cos t + Ce^t $$ The problem asks for the solution to remain finite as \(t \rightarrow \infty\). To satisfy this condition, we need \(Ce^t\) to go to zero as \(t \rightarrow \infty\). Therefore, we must have \(C=0\). Thus, we get the specific solution: $$ y(t) = -1 - 3\cos t $$

Use the initial condition \(y(0) = y_0\): $$ y_0 = -1 - 3\cos (0) $$ Since \(\cos (0) = 1\), we get: $$ y_0 = -1 - 3 \cdot 1 = -4 $$ So the value of \(y_0\) for which the solution remains finite as \(t \rightarrow \infty\) is \(y_0 = -4\).