Question

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

Short answer

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

Step-by-step solution

Rewrite the given equation

Rewrite the given ODE into the standard form for a first-order linear ODE: $$y^{\prime}(t) - 3y(t) = -4$$

Identify the function p(t) and q(t)

In the standard form, the functions p(t) and q(t) are given by: $$p(t) = -3$$ $$q(t) = -4$$

Calculate the integrating factor

The integrating factor is given by the exponential of the integral of p(t) with respect to t: $$\mu(t) = e^{\int p(t) dt} = e^{\int -3 dt} = e^{-3t}$$

Multiply the equation by the integrating factor

Multiply both sides of the equation by the integrating factor, µ(t): $$e^{-3t}(y^{\prime}(t) - 3y(t)) = e^{-3t}(-4)$$

Observe the left side of the equation

Observe that the left side is the derivative of the product of µ(t) and y(t): $$\frac{d}{dt}(e^{-3t}y(t)) = e^{-3t}(-4)$$

Integrate both sides of the equation

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

Solve for y(t)

Divide both sides by the integrating factor, e^{-3t}: $$y(t) =\frac{4}{3} + Ce^{3t}$$

Write the general solution

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

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear ordinary differential equations (ODEs). It transforms a non-exact ODE into an exact one, which can be integrated with ease.

To use the method, you first write the ODE in its standard form: \( y' + p(t)y = q(t) \). The function \( p(t) \) is then utilized to determine the integrating factor \( \mu(t) \), which is typically the exponential of the integral of \( p(t) \), given as \( \mu(t) = e^{\int p(t) dt} \). Multiplying both sides of the ODE by this factor leads to a simplification, enabling us to find a solution.

For instance, consider the equation \( y' - 3y = -4 \). Here, \( p(t) = -3 \) and the integrating factor is \( \mu(t) = e^{\int -3 dt} = e^{-3t} \). By multiplying the entire equation by \( \mu(t) \), we're able to integrate and find the general solution of the ODE.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They come in various types depending on the nature of the derivatives involved. A first-order linear ODE, such as the one presented in the exercise \( y'(t) = 3y(t) - 4 \), involves the first derivative of the function and can describe a wide range of phenomena, from simple growth and decay processes to complex mechanical systems.

Understanding the language of differential equations allows us to model and predict the behavior of dynamic systems in physics, biology, economics, and many other fields. The ability to solve these equations is therefore key in both theoretical studies and practical applications.

General Solution of ODE

The general solution of an ODE encompasses all possible specific solutions and includes arbitrary constants that account for initial conditions. For a first-order ODE, typically there will be a single constant, as it is the case with the provided exercise.

After applying the integrating factor method and simplifying, we obtain an expression that includes an arbitrary constant \( C \). This constant can be determined if an initial condition is provided, leading to a particular solution. In the absence of initial conditions, the solution remains general as in the equation \( y(t) = \frac{4}{3} + Ce^{3t} \), which represents an infinite family of curves, each corresponding to a different value of \( C \).

Exponential Functions

Exponential functions are mathematical functions of the form \( f(t) = a e^{kt} \), where \( a \) and \( k \) are constants, and \( e \) is the base of the natural logarithm. These functions are crucial in solving ODEs because many natural phenomena, like continuous growth or decay, are governed by rates of change proportional to the state of the system.

Exponential functions are unique due to their property of being the derivative of themselves multiplied by a constant. In the context of our exercise, the function \( Ce^{3t} \) is the solution to the homogeneous equation \( y' - 3y = 0 \) and it appears in the general solution to represent all possible behaviors of the system under the influence of the initial value which is represented by \( C \).