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Find the general solution of the first-order, linear equation. $$ y^{\prime}-3 y=5 $$

Short Answer

Expert verified
The general solution is \( y = -\frac{5}{3} + Ce^{3x} \).

Step by step solution

01

Identify the Standard Form

The given differential equation is \( y^{\prime} - 3y = 5 \). This is already in the standard form \( y^{\prime} + P(x)y = Q(x) \) where \( P(x) = -3 \) and \( Q(x) = 5 \).
02

Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x)\,dx} \). Here, \( P(x) = -3 \), so we need to compute the integral:\[\int -3\,dx = -3x\]Therefore, the integrating factor is:\[\mu(x) = e^{-3x}\]
03

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by \( \mu(x) = e^{-3x} \):\[e^{-3x}(y^{\prime} - 3y) = e^{-3x} \cdot 5\]Simplifying, we have:\[(e^{-3x}y)^{\prime} = 5e^{-3x}\]
04

Integrate Both Sides

We integrate both sides of \[(e^{-3x}y)^{\prime} = 5e^{-3x}\]The left side integrates directly to \( e^{-3x}y \), and the right side integrates as follows:\[\int 5e^{-3x}\,dx = \frac{5}{-3}e^{-3x} + C\]Therefore, \[e^{-3x}y = -\frac{5}{3}e^{-3x} + C\]
05

Solve for y

Solve the equation \( e^{-3x}y = -\frac{5}{3}e^{-3x} + C \) for \( y \) by multiplying through by \( e^{3x} \):\[y = -\frac{5}{3} + Ce^{3x}\]This is the general solution to the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Linear Equations
First-order linear equations are a special category of differential equations. They are known for involving the first derivative of the unknown function, often represented as \( y' \). These equations are called 'linear' because the unknown function and its derivative appear only as the first power. No higher degrees or nonlinear operations are involved.
To identify a first-order linear equation, look for the structure:
  • \( y' + P(x)y = Q(x) \),
where \( y' \) is the derivative of \( y \) with respect to \( x \), and \( P(x) \) and \( Q(x) \) are functions of \( x \).
Such equations are widespread across scientific, engineering, and practical applications as they often model growth processes, electrical circuits, or mixing problems.
Integrating Factor
The integrating factor is a crucial tool in solving first-order linear differential equations. It is a specially constructed function that simplifies the differential equation, making it easier to solve.
To find the integrating factor \( \mu(x) \), you calculate:
  • \( \mu(x) = e^{\int P(x) \,dx} \)
Starting with the standard form \( y' + P(x)y = Q(x) \), compute the integral of \( P(x) \) with respect to \( x \). The integrating factor then transforms the equation into an exact derivative on one side, \( (\mu(x)y)' = \mu(x)Q(x) \), which can be directly integrated.
This method saves a lot of algebraic manipulation by automating the process of solving differential equations.
General Solution
The general solution of a differential equation includes all possible solutions, incorporating arbitrary constants that can be adapted to specific conditions. For the first-order linear equation, after applying the integrating factor, integrate both sides to recover \( y \).
The process involves:
  • Transforming the differential equation using the integrating factor,
  • Integrating both sides of the simplified equation,
  • Re-solving for \( y \) if necessary.
In our example, the general solution is \( y = -\frac{5}{3} + Ce^{3x} \), where \( C \) represents the arbitrary constant. This solution allows you to accommodate various initial conditions by adjusting \( C \). Understanding this concept is vital for applying differential equations to real-world scenarios.
Standard Form
The standard form of a first-order linear differential equation is a key element in applying systematic solution strategies. Recognizing the standard form allows us to identify and use appropriate techniques such as integrating factors effectively.
A first-order linear equation in its standard form looks like this:
  • \( y' + P(x)y = Q(x) \)
Here, \( y' \) denotes the derivative of \( y \), while \( P(x) \) and \( Q(x) \) are specific functions of \( x \). The purpose of rewriting an equation into this form is to streamline the solution process. In our example, \( y' - 3y = 5 \) was already suitably arranged, making the subsequent calculation steps, such as finding the integrating factor, straightforward.
Standard forms play a crucial role in organizing and solving differential equations effectively.

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Most popular questions from this chapter

Two tanks, Tank I and Tank II, ate filled with \(V\) gal of pure water. A solution containing \(a \mathrm{lb}\) of salt per gallon of water is poured into Tank I at a rate of \(b\) gal per minute. The solution leaves Tank I at a rate of \(b \mathrm{gal} / \mathrm{min}\) and enters Tank II at the same rate (b gal/min). A drain is adjusted on Tank II and solution leaves Tank II at a rate of \(b \mathrm{gal} / \mathrm{min}\). This keeps the volume of solution constant in both tanks ( \(V \mathrm{gal})\). Show that the amount of salt solution in Tank II, as a function of time \(t\), is given by \(a V-a b t e^{-(b / V) t}-a V e^{-(b / V) t}\).

A lake, with volume \(V=100 \mathrm{~km}^{3}\), is fed by a river at a rate of \(\mathrm{km}^{3} / \mathrm{yr}\). In addition, there is a factory on the lake that introduces a pollutant into the lake at the rate of \(p \mathrm{~km}^{3} / \mathrm{yr}\). There is another river that is fed by the lake at a rate that keeps the volume of the lake constant. This means that the rate of flow from the lake into the outlet river is \((p+r) \mathrm{km}^{3} / \mathrm{yr}\). Let \(x(t)\) denote the volume of the pollutant in the lake at time \(t\). Then \(c(t)=x(t) / V\) is the concentration of the pollutant. (a) Show that, under the assumption of immediate and perfect mixing of the pollutant into the lake water, the concentration satisfies the differential equation $$ c^{\prime}+\frac{p+r}{V} c=\frac{p}{V} . $$ (b) It has been determined that a concentration of over \(2 \%\) is hazardous for the fish in the lake. Suppose that \(r=50 \mathrm{~km}^{3} / \mathrm{yr}_{\mathrm{r}} p=2 \mathrm{~km}^{3} / \mathrm{yr}\), and the initial concentration of pollutant in the lake is zero. How long will it take the lake to become hazardous to the health of the fish?

Determine which of the equations are exact and solve the ones that are. $$ \left(1+\frac{y}{x}\right) d x-\frac{1}{x} d y=0 $$

If the given differential equation is autonomous, identify the equilibrium solution(s). Use a numerical solver to sketch the direction field and superimpose the plot of the equilibrium solution(s) on the direction field. Classify each equilibrium point as either unstable or asymptotically stable. $$ y^{\prime}=1-2 y+y^{2} $$

If the given differential equation is autonomous, identify the equilibrium solution(s). Use a numerical solver to sketch the direction field and superimpose the plot of the equilibrium solution(s) on the direction field. Classify each equilibrium point as either unstable or asymptotically stable. $$ x^{\prime}=t^{2}-x^{2} $$

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