Question

Solve the following differential equations by using integrating factors. $$ y^{\prime}=3 y+2 $$

Short answer

The solution is \( y = -\frac{2}{3} + Ce^{3x} \).

Step-by-step solution

Recognize the Equation Form

The differential equation given is \( y' = 3y + 2 \). This is a first-order linear ordinary differential equation of the form \( y' + P(x)y = Q(x) \). Here we can write it as \( y' - 3y = 2 \), identifying \( P(x) = -3 \) and \( Q(x) = 2 \).

Determine the Integrating Factor

The integrating factor \( \mu(x) \) is found using \( \mu(x) = e^{\int P(x) \, dx} \). Substitute \( P(x) = -3 \):\[\mu(x) = e^{-3x}\].

Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(x) = e^{-3x} \):\[e^{-3x}y' - 3e^{-3x}y = 2e^{-3x}\].

Rewrite the Left Side as a Derivative

The left side of the equation is now the derivative of \( e^{-3x}y \).\[\frac{d}{dx}(e^{-3x}y) = 2e^{-3x}\].

Integrate Both Sides

Integrate both sides with respect to \( x \):\[\int \frac{d}{dx}(e^{-3x}y) \, dx = \int 2e^{-3x} \, dx\].This gives:\[e^{-3x}y = -\frac{2}{3}e^{-3x} + C\], where \( C \) is the integration constant.

Solve for y

Multiply through by \( e^{3x} \) to solve for \( y \):\[y = -\frac{2}{3} + Ce^{3x}\].

Key concepts

First-Order Linear Differential Equations

First-order linear differential equations often appear in mathematics and engineering. A differential equation of this type will have the form \( y' + P(x)y = Q(x) \), where \( y' \) is the derivative of \( y \) with respect to \( x \), \( P(x) \) and \( Q(x) \) are known functions of \( x \). The linearity comes from the fact that the dependent variable \( y \) and its derivative are only being multiplied by functions of \( x \).One key to solving these equations is recognizing them in their standard form and using techniques such as integrating factors to solution them. When the auxiliary function \( P(x) \) or the forcing function \( Q(x) \) changes, the differential nature of the problem changes, but the method of integrating factors remains potent for finding solutions. Breaking down a problem step-by-step, you first rewrite the equation in its canonical form, recognize \( P(x) \) and use it to calculate the integrating factor, \( \mu(x) \). Understanding these equations well provides a powerful tool for addressing more complex scenarios.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve one independent variable and its derivatives. They come in many varieties, including linear and nonlinear, and play a crucial role in modeling real-world phenomena. Each type of ODE, like the first-order linear one considered in this exercise, has specific methods for solving them.The ordinary nature implies that we are interested in functions of a single variable. In our example, we are considering \( y \) as a function of \( x \), with \( y' \) representing the derivative of \( y \). These types of problems can model situations ranging from simple growth processes to complex changes in biological systems or even financial markets.Key to solving these equations, especially linear ones, is transforming the equation into a form where integration becomes straightforward. Techniques like separation of variables, characteristic equations, or, as in our problem, integrating factors, are mainstays in an ODE solver’s toolkit.

Integration Constant

In solving any differential equation, when you integrate, an integration constant \( C \) naturally arises. This constant is important because it embodies the "general solution" to the differential equation, representing a family of curves rather than a single solution.When solving a differential equation, once the left side is transformed and integrated, as was done with \( \frac{d}{dx}(e^{-3x}y) \), integrating both sides of the equation introduces this constant \( C \). It is not just an artifact, but a critical part of the solution that allows for adjustment in initial conditions or boundary values.In physical contexts, \( C \) can often be determined by additional information about the system being modeled, such as initial conditions that specify a starting value. In our exercise, solving for \( y \) ultimately leads to \( y = -\frac{2}{3} + Ce^{3x} \), where adjusting \( C \) gives different solution paths based on starting points or extra conditions applied to the problem.