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Chapter 31: Problem 3

Solve the given differential equations. $$\frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-6 y=0$$

Short Answer

Expert verified
The solution is \( y(x) = C_1 e^{3x} + C_2 e^{-2x} \).

Step by step solution

01

Characteristic Equation

First, we convert the differential equation to its characteristic form. The given equation \( \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 6y = 0 \) becomes \( r^{2} - r - 6 = 0 \).
02

Solve the Quadratic Equation

Next, solve the characteristic equation \( r^{2} - r - 6 = 0 \) to find the roots. This factors into \( (r - 3)(r + 2) = 0 \), giving roots \( r = 3 \) and \( r = -2 \).
03

Formulate the General Solution

Using the roots from Step 2, the general solution of the differential equation is \( y(x) = C_1 e^{3x} + C_2 e^{-2x} \), 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.

Second Order Differential Equations
A second order differential equation involves the second derivative of a function. It typically appears in the form \( a\frac{d^{2} y}{d x^{2}} + b\frac{d y}{d x} + cy = 0 \), where \( a \), \( b \), and \( c \) are constants.
This type of equation describes systems with various applications in physics and engineering, such as oscillations and wave propagation.
In our example, the equation \( \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 6 y = 0 \) is a second order linear differential equation with constant coefficients.
To solve it, we utilize methods involving characteristic equations which simplify the process.
Characteristic Equation
The characteristic equation is derived from a differential equation to simplify finding solutions.
It involves substituting the differential operators with algebraic terms. For instance, in the equation \( \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 6y = 0 \), we transform it to \( r^{2} - r - 6 = 0 \).
This conversion changes the problem to finding roots of a simple quadratic equation instead of solving a differential equation directly.
Solving the characteristic equation gives the values of \( r \), which guide us to the general solution of the differential equation.
General Solution of Differential Equations
The general solution of a differential equation involves finding a formula that describes all possible solutions.
In our case, after obtaining the characteristic roots \( r = 3 \) and \( r = -2 \), we construct the general solution as \( y(x) = C_1 e^{3x} + C_2 e^{-2x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
These constants are determined by initial or boundary conditions, specific to each problem scenario.
The importance of the general solution is that it encapsulates the behavior of the system described by the differential equation, providing insight into the dynamics of the modeled situation.

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Most popular questions from this chapter

The voltage \(v\) at a distance \(s\) along a transmission line is given by \(d^{2} v / d s^{2}=a^{2} v,\) where \(a\) is called the attenuation constant. Solve for \(v\) as a function of \(s\).

Solve the given problems by solving the appropriate differential equation. Assume that the rate of depreciation of an object is proportional to its value at any time \(t .\) If a car costs \(\$ 33,000\) new and its value 3 years later is \(\$ 19,700\) what is its value 11 years after it was purchased?

Solve the given problems by solving the appropriate differential equation. According to Newton's law of cooling, the rate at which a body cools is proportional to the difference in temperature between it and the surrounding medium. Assuming Newton's law holds, how long will it take a cup of hot water, initially at \(200^{\circ} \mathrm{F},\) to cool to \(100^{\circ} \mathrm{F}\) if the room temperature is \(80.0^{\circ} \mathrm{F},\) if it cools to \(140^{\circ} \mathrm{F}\) in 5.0 min?

Find the inverse transforms of the given functions of \(s\). $$F(s)=\frac{3 s^{4}+3 s^{3}+6 s^{2}+s+1}{s^{5}+s^{3}}$$ (Explain your method of solution.)

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. $$y^{*}+y=1, y(0)=1, y^{\prime}(0)=1$$
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