Question

What is the general solution of the equation \(y^{\prime}(t)=-4 y+6 ?\)

Short answer

Answer: The general solution of the given ODE is \(y(t)= \frac{3}{2} + Ce^{-4t}\), where \(C\) is the constant of integration.

Step-by-step solution

Identify the integrating factor

The given ODE can be expressed as \(\frac{dy}{dt} + 4y = 6\). In this case, the function \(P(t)\) is constant and equal to \(4\). Therefore, the integrating factor is given by \(I(t) = e^{\int P(t) dt}\). Let's calculate the integrating factor: \(I(t) = e^{\int 4 dt} = e^{4t}\)

Multiply the whole equation by the integrating factor

Now, we will multiply the entire equation by the integrating factor \(I(t) = e^{4t}\): \(e^{4t} \frac{dy}{dt} + 4e^{4t}y = 6e^{4t}\)

Integrate both sides of the equation

Observe that the left side of the resulting equation is the derivative of the product of \(y(t)\) and the integrating factor \(I(t)\), i.e., \(\frac{d}{dt} (ye^{4t})\). Now, we will integrate both sides of the equation with respect to \(t\): \(\int \frac{d}{dt}(ye^{4t}) dt = \int 6e^{4t} dt\) Integrating both sides, we obtain: \(ye^{4t} = \frac{3}{2}e^{4t} + C\), where \(C\) is the constant of integration.

Write the general solution

Now, we will isolate \(y(t)\): \(y(t) = \frac{3}{2} + Ce^{-4t}\) Hence, the general solution of the given equation is \(y(t) = \frac{3}{2} + Ce^{-4t}\).