Question

What is the general solution of the equation \(y^{\prime}(t)=3 y-12 ?\)

Short answer

Question: Determine the general solution of the first-order linear differential equation \(y'(t) = 3y - 12\). Answer: The general solution of the given differential equation is \(y(t) = -4 + Ce^{3t}\), where \(C\) is an arbitrary constant.

Step-by-step solution

Rewrite the equation in standard form

To solve the given differential equation, we first rewrite it as a first-order linear differential equation. In standard form, we have: $$y'(t) - 3y = -12$$ This is now a first-order linear differential equation with \(P(t) = -3\) and \(Q(t) = -12\).

Find the integrating factor

To find the general solution to this equation, we will use an integrating factor. The integrating factor is given by \(e^{\int P(t) dt}\). In this case, our integrating factor is: $$e^{\int -3 dt} = e^{-3t}$$

Multiply the differential equation by the integrating factor

Now we multiply the given differential equation by the integrating factor: $$e^{-3t}(y'(t) - 3y) = -12e^{-3t}$$ This simplifies the left side of the equation to the derivative of \((ye^{-3t})\) with respect to \(t\): $$(ye^{-3t})' = -12e^{-3t}$$

Integrate both sides of the equation

Now we can integrate both sides of the equation to solve for \(y\): $$\int (ye^{-3t})' dt = \int -12e^{-3t} dt $$ On the left, the integral of the derivative is simply the function itself: $$ye^{-3t} = \frac{-4e^{-3t}}{1} + C$$ Now, to find the general solution, we can multiply both sides by \(e^{3t}\) to isolate \(y\):

Isolate y and write the general solution

Multiply both sides by \(e^{3t}\): $$y(t) = \frac{-4e^{-3t}}{1}e^{3t} + Ce^{3t}$$ This simplifies to: $$y(t) = -4 + Ce^{3t}$$ This is the general solution of the given differential equation. Here, \(C\) is an arbitrary constant which can be found if an initial condition is given.