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Chapter 8: Problem 7

Solve the following differential equations. \(\left(D^{2}-5 D+6\right) y=0\)

Short Answer

Expert verified
The general solution is \(y = C_{1} e^{2x} + C_{2} e^{3x}\)

Step by step solution

01

Identify the Characteristic Equation

Convert the differential equation to its characteristic equation by replacing \(D\) with \(r\). This yields the characteristic equation: \(r^{2} - 5r + 6 = 0\)
02

Solve the Characteristic Equation

Solve the quadratic characteristic equation \(r^{2} - 5r + 6 = 0\) using the factored form or the quadratic formula. Factoring the equation, we get: \((r - 2)(r - 3) = 0\)
03

Find the Roots

Set each factor equal to zero to find the roots of the characteristic equation. This yields \(r = 2\) and \(r = 3\)
04

Construct the General Solution

With the roots of the characteristic equation being real and distinct, the general solution to the differential equation is: \(y = C_{1} e^{2x} + C_{2} e^{3x}\) 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.

Understanding the Characteristic Equation
In the context of solving differential equations, the first step is often finding the characteristic equation. This beautiful tool helps simplify complex differential equations to more manageable algebraic ones.
To convert a differential equation like \((D^{2} - 5D + 6)y = 0\), replace the differentiation operator \(D\) with a variable, typically denoted as \(r\).

Doing this for our example, we get the characteristic equation as:
  • \(r^{2} - 5r + 6 = 0\)

This quadratic equation is simpler to handle compared to the original differential form.
Mastering the Quadratic Formula
Solving the characteristic equation often means using the quadratic formula, especially when factoring is not straightforward. The quadratic formula provides roots for any quadratic equation of the form \(ax^{2} + bx + c = 0\):

\[ r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
For our characteristic equation \(r^{2} - 5r + 6 = 0\), here are the coefficients:
  • \(a = 1\)
  • \(b = -5\)
  • \(c = 6\)

We apply these values to the quadratic formula and solve for \(r\):

\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4*1*6}}{2*1} \]
Upon computing, we get \(r = 2\) and \(r = 3\). These are the roots of the equation.
Formulating the General Solution
With the roots identified, the last step is constructing the general solution to our differential equation. The general solution is a combination of exponential functions based on the roots:
  • For roots that are real and distinct, say \(r_{1}\) and \(r_{2}\), the general solution is:
    \(y = C_{1}e^{r_{1}x} + C_{2}e^{r_{2}x}\).

  • \(C_{1}\) and \(C_{2}\) are arbitrary constants determined by initial conditions or boundary conditions provided.
In our example, with roots \(r_{1} = 2\) and \(r_{2} = 3\), the general solution becomes:
\(y = C_{1}e^{2x} + C_{2}e^{3x}\).

This structure ensures we have a versatile solution that can fit a variety of scenarios by adjusting constants \(C_{1}\) and \(C_{2}\).

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Most popular questions from this chapter

Find the general solution of each of the following differential equations. \(\frac{d x}{d y}=\cos y-x \tan y\)

The differential equation for the path of a planet around the sun (or any object in an inverse square force field) is, in polar coordinates, $$ \frac{1}{r^{2}} \frac{d}{d \theta}\left(\frac{1}{r^{2}} \frac{d r}{d \theta}\right)-\frac{1}{r^{3}}=-\frac{k}{r^{2}} $$ Make the substitution \(u=1 / r\) and solve the equation to show that the path is a conic section.

Verify that \(y=\sin x, y=\cos x, y=e^{i x}\), and \(y=e^{-i x}\) are all solutions of \(y^{\prime \prime}=-y\).

Find the general solution of each of the following differential equations. \(\frac{d y}{d x}=\frac{3 y}{3 y^{2 / 3}-x}\)

The velocity of a particle on the \(x\) axis, \(x \geq 0\), is always numerically equal to the square root of its displacement \(x\). If \(v=0\) when \(x=0\), find \(x\) as a function of \(t .\) Show that the given conditions are satisfied if the particle remains at the origin for any arbitrary length of time \(t_{0}\), and then moves away; find \(x\) for \(t>t_{0}\) for this case.
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