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Chapter 9: Problem 1

Solve. \(y^{\prime \prime}-6 y^{\prime}+5 y=0\)

Short Answer

Expert verified
The general solution is \(y(t) = C_1 e^{5t} + C_2 e^{t}\).

Step by step solution

01

- Form the Characteristic Equation

To solve the differential equation, first form the characteristic equation by assuming a solution of the form \(y = e^{rt}\). Plugging \(y = e^{rt}\), \(y' = re^{rt}\), and \(y'' = r^2e^{rt}\) into the differential equation results in the characteristic equation: \(r^2 - 6r + 5 = 0\).
02

- Solve the Characteristic Equation

Solve the quadratic characteristic equation \(r^2 - 6r + 5 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 5\). This gives: \[r = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2}\] Thus, \(r = 5\) and \(r = 1\).
03

- Form the General Solution

Using the roots from the characteristic equation, the general solution to the differential equation is formed. For roots \(r_1\) and \(r_2\), the solution is: \[y(t) = C_1 e^{r_1t} + C_2 e^{r_2t}\]Substituting \(r_1 = 5\) and \(r_2 = 1\), the general solution is: \[y(t) = C_1 e^{5t} + C_2 e^{t}\]

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Key Concepts

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

characteristic equation
The characteristic equation is crucial when solving a second-order linear homogeneous differential equation like the one in our exercise: \( y'' - 6y' + 5y = 0 \). To form this equation, we assume a solution in the form of \( y = e^{rt} \). By substituting \( y \), \( y' \), and \( y'' \) into the original differential equation, we derive the characteristic equation: \( r^2 - 6r + 5 = 0 \). This characteristic equation simplifies our problem into solving for \( r \), the roots of the equation, which will guide us in forming the general solution.
quadratic formula
The quadratic formula is a straightforward method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is given by: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our characteristic equation \( r^2 - 6r + 5 = 0 \), we have \( a = 1 \), \( b = -6 \), and \( c = 5 \). Plugging these values into the quadratic formula gives us: \[ r = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2} \] Solving this, we get two distinct roots: \( r = 5 \) and \( r = 1 \). These roots provide us with the necessary components to write the general solution of the differential equation.
general solution
The general solution of a second-order linear homogeneous differential equation is formed using the roots obtained from the characteristic equation. In our case, with roots \( r = 5 \) and \( r = 1 \), the general solution is: \[ y(t) = C_1 e^{5t} + C_2 e^{t} \] Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions or boundary conditions provided in the problem. This form of the solution captures all possible solutions to the differential equation, expressing them in terms of exponential functions dictated by the characteristic roots.

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Most popular questions from this chapter

Solve the initial-value problem. \(y^{\prime \prime}+2 y^{\prime}+2 y=2, y(0)=2, y^{\prime}(0)=1\)

Solve the initial-value problem. \(x^{\prime}=2 x+3 y, y^{\prime}=-3 x+8 y, x(0)=1, y(0)=2\)

The matrix method may also be used for systems of three or more functions. For Exercises \(44-49\), find the general solution. \(x^{\prime}=3 x+2 z, y^{\prime}=-x+2 y-2 z, z^{\prime}=-x\)

Solve the initial-value problem. \(y^{\prime \prime}-y^{\prime}-2 y=2 x-1, y(0)=6, y^{\prime}(0)=0\)

Determine the eigenvalues for the system of differential equations. If the eigenvalues are real and distinct, find the general solution by determining the associated eigenvectors. If the eigenvalues are complex or repeated, solve using the reduction method. \(x^{\prime}=-2 x-2 y, y^{\prime}=x\)
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