Question

Solve the following initial value problems. $$y^{\prime}(t)-2 y=8, y(0)=0$$

Short answer

Question: Determine the particular solution for the given initial value problem involving a first-order linear differential equation: $y'(t) - 2y(t) = 8$, with an initial condition $y(0) = 0$. Answer: The particular solution for the given initial value problem is $y(t) = -4 + 4e^{2t}$.

Step-by-step solution

Identify the type of differential equation

This is a first-order linear differential equation in the form: $$y'(t) + P(t)y(t) = Q(t)$$ Where \(P(t) = -2\) and \(Q(t) = 8\).

Find the integrating factor

We need to find the integrating factor, which is given by: $$\mu(t) = e^{\int P(t) dt}$$ In our case, \(P(t) = -2\), so we have: $$\mu(t) = e^{\int -2 dt} = e^{-2t}$$

Multiply both sides by the integrating factor

Multiply both sides of the equation by the integrating factor \(\mu(t) = e^{-2t}\): $$(e^{-2t}y'(t)) - 2e^{-2t}y(t) = 8e^{-2t}$$ The left side of this equation is now the derivative of the product \(\mu(t)y(t)\): $$\frac{d}{dt}(e^{-2t}y(t)) = 8e^{-2t}$$

Integrate both sides with respect to t

Now, we integrate both sides with respect to \(t\): $$\int\frac{d}{dt}(e^{-2t}y(t)) dt = \int 8e^{-2t} dt$$ This yields: $$e^{-2t}y(t) = -4e^{-2t} + C$$

Solve for y(t)

Now, we solve for \(y(t)\) by multiplying both sides by \(e^{2t}\): $$y(t) = -4 + Ce^{2t}$$

Apply the initial condition to find the particular solution

Finally, to find the particular solution, substitute the initial condition \(y(0) = 0\) into the form of the general solution: $$0 = -4 + Ce^{2 \times 0} \Rightarrow 0 = -4 + C$$ This gives \(C = 4\), so the particular solution is: $$y(t) = -4 + 4e^{2t}$$