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Chapter 4: Problem 15

\(y^{\prime \prime}-6 y^{\prime}+9 y=0\)

Short Answer

Expert verified
The general solution is y(t) = (C1 + C2t) e^(3t).

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^{u}-6 y^{u}+9 y=0
02

Write the Characteristic Equation

To solve this type of differential equation, first, find the characteristic equation by replacing y^{u} with r^{2} y^{u} and y^{u} with r. The characteristic equation for the given differential equation is:
03

Solve the Characteristic Equation

Solve the characteristic equation: r^{2}-6r+9=0. This factors to:
04

Find the Roots of the Characteristic Equation

Factor the quadratic equation to find its roots: (r-3)^{2}=0. Therefore, the root is r=3, which is a repeated root.
05

Write the General Solution

Since the characteristic equation has a repeated root, the general solution to the differential equation is: y(t) = (C_{1} + C_{2}t) e^{3t} .

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Key Concepts

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

characteristic equation
When solving a second-order linear homogeneous differential equation like \( y^{\text{''}} -6 y^{\text{'} } + 9 y = 0 \), one of the initial steps is to write the **characteristic equation**. This equation helps us determine the form of the solution. To form the characteristic equation:
  • Replace \( y^{\text{''}} \) with \( r^2 \)
  • Replace \( y^{\text{'}} \) with \( r \)
Doing so, transforms the differential equation into a quadratic equation: \[ r^2 - 6r + 9 = 0 \]This is the characteristic equation. Solving this quadratic equation will provide the roots, which are crucial for finding the general solution.
repeated roots
After forming the characteristic equation, the next step is to solve it. In our example, the characteristic equation is \[ r^2 - 6r + 9 = 0 \]Solving this, we factor it as \[ (r-3)^2 = 0 \]This indicates that there is only one root, \( r = 3 \), but it has a multiplicity of 2 (it is a repeated root). Repeated roots occur when the same root appears more than once. In such cases, the solutions are not simple exponentials but involve multiples of \( t \). Understanding this is crucial for forming the general solution.
general solution
The general solution for a second-order linear homogeneous differential equation depends on the roots of its characteristic equation. For repeated roots, like in the given problem where \( r = 3 \) with multiplicity 2, the general solution takes a specific form to account for the repetition. If \( r_1 \) is a repeated root, the general solution is \[ y(t) = (C_1 + C_2t) e^{r_1 t} \]In our specific case: \[ y(t) = (C_1 + C_2t) e^{3t} \]This formula ensures that the two constants \( C_1 \) and \( C_2 \) account for different initial conditions or boundary conditions. These constants are found through additional information, often given in the problem.

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Most popular questions from this chapter

\(y^{\prime \prime \prime}+10 y^{\prime \prime}+34 y^{\prime}+40 y=t e^{-4 t}+2 e^{-3 t} \cos t\)

(a) Solve the initial value problem \(6 x^{2} y^{\prime \prime}+\) \(5 x y^{\prime}-y=0, y(1)=a, y^{\prime}(1)=b\). (b) Find conditions on \(a\) and \(b\) so that \(\lim _{x \rightarrow 0^{+}} y(x)=0\). Graph several solutions to confirm your results. (c) Find conditions on \(a\) and \(b\) so that \(\lim _{x \rightarrow \infty} y(x)=0\). Graph several solutions to confirm your results. (d) If both \(a\) and \(b\) are not zero, is it possible to find \(a\) and \(b\) so that both \(\lim _{x \rightarrow 0^{+}} y(x)=0\) and \(\lim _{x \rightarrow \infty} y(x)=0\) ? Explain.

\(2 x^{2} y^{\prime \prime}+3 x y^{\prime}-y=x^{-2}, y(1)=0, y^{\prime}(1)=2\)

\(4 x^{2} y^{\prime \prime}+8 x y^{\prime}+y=0\)

Solve each of the following initial value problems. Verify that your result satisfies the initial conditions by graphing it on an appropriate interval. (a) \(y^{\prime \prime \prime}+3 y^{\prime \prime}+2 y^{\prime}+6 y=0, y(0)=0, y^{\prime}(0)=\) \(1, y^{\prime \prime}(0)=-1\) (b) \(y^{(4)}-8 y^{\prime \prime \prime}+30 y^{\prime \prime}-56 y^{\prime}+49 y=0\), \(y(0)=1, y^{\prime}(0)=2, y^{\prime \prime}(0)=-1, y^{\prime \prime \prime}(0)=\) \(-1\) (c) \(0.31 y^{\prime \prime \prime}+11.2 y^{\prime \prime}-9.8 y^{\prime}+5.3 y=0\), \(y(0)=-1, y^{\prime}(0)=-1, y^{\prime \prime}(0)=0\)
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