Problem 21 - Calculus Early Transcendentals

Question

Find the general solution of the following equations. $$y^{\prime}(t)=3 y-4$$

Step-by-Step Solution

Answer

Answer: The general solution of the given ODE is $y(t) = \frac{4}{3} + Ce^{3t}$.

Step 1: Identify the differential equation type

Determine whether the given differential equation is linear or not. In this case, the given ODE is a first-order linear ODE: $$y^{\prime}(t)=3 y-4$$

Step 2: Write the equation in standard form

Write the given ODE in its standard linear form: $$y^{\prime}(t) -3y = -4$$

Step 3: Find the integrating factor

Calculate the integrating factor: $$\mu(t) = e^{\int{-3\, dt}} = e^{-3t}$$

Step 4: Multiply the differential equation by the integrating factor

Multiply the standard form ODE by the integrating factor: $$e^{-3t}(y^{\prime}(t) -3y) = e^{-3t}(-4)$$

Step 5: Check that the left-hand side is an exact derivative

Check if the left-hand side is an exact derivative, and in this case, it is: $$\frac{d}{dt}(e^{-3t}y(t)) = e^{-3t}(y^{\prime}(t) -3y)$$

Step 6: Integrate both sides with respect to t

Integrate both sides of the equation with respect to t: $$\int \frac{d}{dt}(e^{-3t}y(t)) \, dt = \int e^{-3t}(-4) \, dt$$

Step 7: Solve the integral

Solving the integral, we get: $$e^{-3t}y(t) = \frac{4}{3}e^{-3t} + C$$

Step 8: Isolate y(t) to find the general solution

Finally, isolate y(t) to find the general solution of the ODE: $$y(t) = \frac{4}{3} + Ce^{3t}$$