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Find the general solution to the given differential equation. $$y^{\prime \prime}+4 y^{\prime}+3 y=0$$

Short Answer

Expert verified
The general solution is \( y(t) = C_1 e^{-3t} + C_2 e^{-t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is a second-order linear homogeneous differential equation with constant coefficients: \[ y'' + 4y' + 3y = 0 \]
02

Write the Characteristic Equation

To find the general solution, we first write the characteristic equation of the form \( ar^2 + br + c = 0 \). For the given differential equation, replace \(a\) with 1, \(b\) with 4, and \(c\) with 3. Thus, the characteristic equation is:\[ r^2 + 4r + 3 = 0 \]
03

Solve the Characteristic Equation

Factor the quadratic characteristic equation:\[ (r + 3)(r + 1) = 0 \]Thus, the roots of the characteristic equation are \( r_1 = -3 \) and \( r_2 = -1 \).
04

Write the General Solution

Since the roots \( r_1 \) and \( r_2 \) are real and distinct, the general solution to the differential equation is given by:\[ y(t) = C_1 e^{-3t} + C_2 e^{-t} \]where \( C_1 \) and \( C_2 \) are arbitrary constants.

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Key Concepts

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

Characteristic Equation
In solving second-order linear differential equations, one of the primary steps is finding the characteristic equation. This equation is derived from the original differential equation by replacing the derivatives with terms involving a variable, often denoted by \( r \). For a general second-order differential equation of the form:
  • \( y'' + by' + cy = 0 \)
We write its characteristic equation as:
  • \( ar^2 + br + c = 0 \)
This quadratic equation allows us to determine the nature of the solutions by finding its roots. These roots, \( r_1 \) and \( r_2 \), can be real and distinct, repeated, or complex, which further dictate the form of the general solution to the differential equation.
In our example, the given differential equation \( y'' + 4y' + 3y = 0 \) leads to the characteristic equation \( r^2 + 4r + 3 = 0 \). Solving this equation by factoring, we find that the roots are \( r_1 = -3 \) and \( r_2 = -1 \). These roots being real and distinct, play a crucial role in constructing the general solution.
Second-order Linear Differential Equations
Second-order linear differential equations are a type of differential equation characterized by the second highest derivative of an unknown function. The general form of these equations is:
  • \( a(t)y'' + b(t)y' + c(t)y = g(t) \)
where \( a(t), b(t), c(t), \) and \( g(t) \) are given functions of \( t \), and \( y'' \) represents the second derivative of \( y \).
When the functions \( a(t), b(t), \) and \( c(t) \) are constants, the equation becomes linear with constant coefficients, as seen in the example \( y'' + 4y' + 3y = 0 \).
The hallmark of these equations is that they often require characteristic equations to solve. Depending on the nature of the roots of these characteristic equations, we can determine the form of the particular solution which involves exponential, trigonometric, or even polynomial functions. Second-order linear differential equations are pivotal in modeling physical processes like motion, heat conduction, and wave propagation.
Homogeneous Differential Equations
Homogeneous differential equations refer to differential equations where the function \( g(t) = 0 \) in the general form \( a(t)y'' + b(t)y' + c(t)y = g(t) \). This means the equation has no external forcing function or non-homogeneous term.
The homogeneous nature of such equations implies that all terms are dependent on the function \( y \) and its derivatives. The solution to a homogeneous differential equation is typically considered the 'complementary' solution to its non-homogeneous counterpart, if any. For a homogeneous differential equation of the form \( y'' + 4y' + 3y = 0 \), the focus is solely on the behavior dictated by its coefficients.
Solving these equations involves finding the roots of the characteristic equation derived from the differential equation. Each type of root (real, repeated, or complex) has an associated solution form, involving constants that will be determined by initial conditions or boundary values provided. In our example, with distinct real roots, the general solution is \( y(t) = C_1 e^{-3t} + C_2 e^{-t} \), fully determined by the roots of the characteristic equation.

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Most popular questions from this chapter

An item is initially sold at a price of \(\$ p\) per unit. Over time, market forces push the price toward the equilibrium price, \(\$ p^{*},\) at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, \(\$ p,\) is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for \(p\) as a function of \(t.\) (b) Solve for \(p.\) (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to \(p\) as \(t \rightarrow \infty ?\)

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(\lim _{x \rightarrow \infty} g(x)=0,\) then \(\lim _{x \rightarrow \infty} f(x)=0.\)

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Each of the differential equations (i)-(iv) represents the position of a 1 gram mass oscillating on the end of a damped spring. Pick the differential equation representing the system which answers the question. (i) \(\quad s^{\prime \prime}+s^{\prime}+4 s=0\) (ii) \(s^{\prime \prime}+2 s^{\prime}+5 s=0\) (iii) \(s^{\prime \prime}+3 s^{\prime}+3 s=0\) (iv) \(\quad s^{\prime \prime}+0.5 s^{\prime}+2 s=0\) Which oscillation has the longest period?

Find the general solution to the given differential equation. $$\frac{d^{2} p}{d t^{2}}+\frac{d p}{d t}+p=0$$

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