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Find the general solution to the given differential equation. $$y^{\prime \prime}-3 y^{\prime}+2 y=0$$

Short Answer

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

Step by step solution

01

Identify the type of differential equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. It is in the form: \( a y'' + b y' + c y =0 \). In our case, \( a = 1 \), \( b = -3 \), \( c = 2 \).
02

Write the characteristic equation

For a second-order linear homogeneous differential equation \( y'' + b y' + c y = 0 \), the characteristic equation is obtained by replacing \( y \) with \( e^{rx} \), which simplifies to \( ar^2 + br + c = 0 \). For the given equation, the characteristic equation is \( r^2 - 3r + 2 = 0 \).
03

Factor the characteristic equation

To find the roots of the characteristic equation \( r^2 - 3r + 2 = 0 \), we can factor it. It factors as \( (r - 1)(r - 2) = 0 \).
04

Solve for the roots of the factorized equation

Set each factor of the characteristic equation to zero to find the roots. The roots are \( r_1 = 1 \) and \( r_2 = 2 \).
05

Write the general solution

Because we have distinct real roots \( r_1 = 1 \) and \( r_2 = 2 \), the general solution to the differential equation is given by \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). Substituting the values, the general solution is \( y(x) = C_1 e^x + C_2 e^{2x} \), 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 equation
A second-order differential equation involves the second derivative of an unknown function. In simpler terms, it describes a situation where the rate of change of a rate of change affects something. In comparison to first-order differential equations, these equations capture more complexity and can describe more nuanced dynamic systems.
When an equation is labeled second-order, it means it includes a term involving the second derivative, denoted as \( y'' \). For example, in the equation \( y'' - 3y' + 2y = 0 \), the term \( y'' \) is the second derivative of \( y \). This indicates that how fast the rate of change of \( y \) itself is changing impacts the situation being modeled. Knowing this, we can understand how such equations describe oscillations, growth, and other physical processes.
Characteristic equation
The characteristic equation is a crucial tool when solving linear differential equations. It's formed during the process of finding functions that satisfy the differential equation. To derive the characteristic equation, we assume solutions have the form \( y = e^{rx} \), where \( r \) is a constant. By substituting this form into the differential equation, we transform the equation into a polynomial in \( r \).
This polynomial, known as the characteristic equation, reflects the nature of solutions to the differential equation. For the given differential equation \( y'' - 3y' + 2y = 0 \), the characteristic equation is \( r^2 - 3r + 2 = 0 \). Solving this polynomial helps identify the roots, which are directly tied to the behavior of the solution.
Linear homogeneous differential equation
Linear homogeneous differential equations are a type of equation where each term is a derivative of the unknown function and the equation equals zero. The absence of standalone functions (non-zero terms on the other side of the equation) means solutions can be added together to form new solutions, emphasizing linearity.
In the context of our example, \( y'' - 3y' + 2y = 0 \) is linear since it only involves terms of \( y \) and its derivatives. It's also homogeneous because the equation totals to zero. This structure allows for solutions to be expressed in terms of arbitrary constants, reflecting the infinite set of solutions that satisfy the equation under varying initial conditions.
Constant coefficients
A differential equation with constant coefficients is one where the coefficients are not variables but constants. This characteristic makes the equation easier to analyze and solve, as these coefficients don't change with respect to the independent variable.
For the equation \( y'' - 3y' + 2y = 0 \), the coefficients are \( 1 \), \( -3 \), and \( 2 \). Since these coefficients remain constant, the process of forming the characteristic equation and solving it is straightforward. The constant values ensure that the solution methods remain uniform across different conditions, making such equations a staple in solving more complex systems efficiently.

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Most popular questions from this chapter

Are the statements true or false? Give an explanation for your answer. The system of differential equations \(d x / d t=-x+x y^{2}\) and \(d y / d t=y-x^{2} y\) requires initial conditions for both \(x(0)\) and \(y(0)\) to determine a unique solution.

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 \(\lim _{x \rightarrow \infty} f(x)=0.\)

Two companies share the market for a new technology. They have no competition except each other. Let \(A(t)\) be the net worth of one company and \(B(t)\) the net worth of the other at time \(t .\) Suppose that net worth cannot be negative and that \(A\) and \(B\) satisfy the differential equations $$\begin{aligned}&A^{\prime}=2 A-A B\\\&B^{\prime}=B-A B\end{aligned}$$ (a) What do these equations predict about the net worth of each company if the other were not present? What effect do the companies have on each other? (b) Are there any equilibrium points? If so, what are they? (c) Sketch a slope field for these equations (using a computer or calculator), and hence describe the different possible long-run behaviors.

Give an example of: A second-order differential equation which has \(y=e^{2 x}\) as a solution.

If \(y=e^{2 t}\) is a solution to the differential equation $$ \frac{d^{2} y}{d t^{2}}-5 \frac{d y}{d t}+k y=0 $$ find the value of the constant \(k\) and the general solution to this equation.

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