91Ó°ÊÓ

Let's start by substituting \(z=y+1\) into the given differential equation: \(y^{\prime}=-(y+1)+t(y+1)^{-2}\). We will also need to find \(y^{\prime}\) in terms of \(z\). Since \(z=y+1\), we can find \(y^{\prime}\) by differentiating \(z\) with respect to \(t\): \(z^{\prime}=y^{\prime}\). Now, substitute \(y^{\prime}=z^{\prime}\), \(y=z-1\), and \(y+1=z\) into the given differential equation: \(z^{\prime}=-(z)+t(z)^{-2}\).

Solve the differential equation for \(z\): \(z^{\prime}=-(z)+t(z)^{-2}\). To make this clearer, let's rewrite it as: \(z^{\prime}+z=tz^{-2}\). This is a first-order linear differential equation, which can be solved using an integrating factor. The integrating factor is given by \(e^{\int p(t)dt}\), where \(p(t)=1\). Therefore, the integrating factor is \(e^{\int 1 dt}=e^t\). Now, multiply the differential equation by the integrating factor: \(e^tz^{\prime}+e^tz=te^{t}z^{-2}\). The left-hand side of this equation is the derivative of the product \((e^tz)\) with respect to \(t\). So, we have: \(\frac{d}{dt}(e^tz)=te^{t}z^{-2}\). Now integrate both sides with respect to \(t\): \(\int \frac{d}{dt}(e^tz)dt = \int te^{t}z^{-2}dt\). By using substitution, it becomes easier to integrate the right side: Let \(u=e^t\). Then, \(dt=\frac{du}{u}\). Therefore, \(te^{t}z^{-2}dt=tudu^{-1}z^{-2}\). Now integrate: \(e^tz=\int tudu^{-1}z^{-2}+C\). Now solve it for \(z\).

From the previous step, we have the equation: \(e^tz=\int tudu^{-1}z^{-2}+C\). We need to find the initial condition for \(z\): \(z(0)=y(0)+1=1+1=2\). To find the constant of integration \(C\), let \(z=2\) when \(t=0\): \(2e^0=C\). So, \(C=2\). Therefore, our equation becomes: \(e^tz=\int tudu^{-1}z^{-2}+2\). Now, multiply both sides by \(e^{-t}\): \(z=e^{-t}(\int tudu^{-1}z^{-2}+2)\). We now have the equation for \(z\) in terms of \(t\).

Since we know that \(z=y+1\), we can find \(y\) by rearranging the equation and substituting the expression we found for \(z\): \(y=z-1=e^{-t}(\int tudu^{-1}z^{-2}+2)-1\). We now have an expression for \(y\) in terms of \(t\).