91Ó°ÊÓ

The given differential equation is: $$z'(t)+\frac{z(t)}{2}=6$$

To solve the given linear first-order differential equation, we can use an integrating factor. The integrating factor, denoted by \(I(t)\), can be determined by the formula: $$I(t)=e^{\int p(t) dt}$$ where \(p(t)\) is the coefficient of the z(t) term in the given differential equation. In our case, \(p(t) = \frac{1}{2}\). Thus, the integrating factor becomes: $$I(t)=e^{\int \frac{1}{2} dt}=e^{\frac{t}{2}}$$ Now, multiply the given differential equation by the integrating factor: $$e^{\frac{t}{2}}z'(t)+\frac{e^{\frac{t}{2}}z(t)}{2}=6e^{\frac{t}{2}}$$ Observe that the left-hand side of the resulting equation is the derivative of the product of \(z(t)\) and \(I(t)\). Hence, the above equation can be written as: $$(e^{\frac{t}{2}}z(t))'=6e^{\frac{t}{2}}$$ Integrating both sides with respect to t: $$\int(e^{\frac{t}{2}}z(t))' dt = \int 6e^{\frac{t}{2}} dt$$ Now integrate both sides: $$e^{\frac{t}{2}}z(t)=12e^{\frac{t}{2}}+C$$ where C is the constant of integration. Divide both sides by \(e^{\frac{t}{2}}\), we have: $$z(t)=12+C e^{-\frac{t}{2}}$$

Given that \(z(-1) = 0\), we can substitute this into our expression for \(z(t)\): $$0=12+C e^{\frac{1}{2}}$$ Solving for C, we find that: $$C=-12 e^{\frac{1}{2}}$$

Substituting the value of C back into our expression for \(z(t)\), we have the solution to the initial value problem: $$z(t)=12-12 e^{\frac{1-2t}{2}}$$