Question

Solve the following initial value problems. $$y^{\prime}(t)=3 y-6, y(0)=9$$

Short answer

Question: Determine the function y(t) that satisfies the initial value problem for the given first-order differential equation and initial condition: $\frac{dy}{dt} = 3y - 6$, $y(0) = 9$. Answer: The function y(t) that satisfies the initial value problem is $y(t) = y - 2 + 11e^{-3t}$.

Step-by-step solution

Rewrite the Given Equation

Rewrite the given equation to emphasize that it is a first-order differential equation: $$\frac{dy}{dt} = 3y - 6$$

Solve the Differential Equation

Apply the Integrating Factor method to solve the differential equation. Define the integrating factor, IF, as: $$ IF = e^{\int P(t) dt }$$ Where P(t) is the coefficient of y in the given equation. In our case, P(t) = 3. Calculate the integral of P(t): $$\int P(t) dt = \int 3 dt = 3t$$ Find the integrating factor: $$IF = e^{3t}$$ Multiply both sides of the given differential equation by the integrating factor: $$e^{3t}\frac{dy}{dt} = 3e^{3t}y - 6e^{3t}$$ The left side of the equation is now an exact differential: $$\frac{d}{dt}(e^{3t}y) = 3e^{3t}y - 6e^{3t}$$ Integrate both sides with respect to t: $$\int \frac{d}{dt}(e^{3t}y)dt = \int (3e^{3t}y - 6e^{3t})dt$$ $$e^{3t}y = e^{3t}y - 2e^{3t} + C$$

Solve for y(t)

To find the general solution, solve for y(t): $$y(t) = e^{-3t}(e^{3t}y - 2e^{3t} + C)$$ $$y(t) = y - 2 + Ce^{-3t}$$

Apply the Initial Condition

Now, use the initial condition y(0) = 9 to find the value of the constant C: $$9 = y(0) = y - 2 + Ce^{-3*0}$$ $$9 = y - 2 + C$$ $$C = 11$$

Write the Particular Solution

Substitute the value of C back into the general solution to find the particular solution for the given initial value problem: $$y(t) = y - 2 + 11e^{-3t}$$ The solution to the initial value problem is: $$y(t) = y - 2 + 11e^{-3t}$$