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Chapter 9: Problem 28

Solve. \(y^{\prime \prime}-2 y^{\prime \prime}+4 y^{\prime}-8 y=0\)

Short Answer

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

Step by step solution

01

Identify and Simplify the Differential Equation

The given differential equation simplifies as follows:\( y^{\text{''}} - 2y^{\text{''}} + 4y^{\text{'}} - 8y =0 \)Combine like terms to simplify the equation:\( -y^{\text{''}} + 4y^{\text{'}} - 8y = 0 \) can be rearranged to\( y^{\text{''}} - 4y^{\text{'}} + 8y = 0 \)
02

Set Up the Characteristic Equation

Convert the simplified differential equation to its characteristic equation using the form \( ar^2 + br + c = 0 \).For \( y^{\text{''}} - 4y^{\text{'}} + 8y = 0 \), the characteristic equation is:\( r^2 - 4r + 8 = 0 \).
03

Solve the Characteristic Equation

Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( r \). In this equation, \( a = 1 \), \( b = -4 \), and \( c = 8 \).\( r = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(8)}}{2(1)} = \frac{4 \pm \sqrt{16 - 32}}{2} = \frac{4 \pm \sqrt{-16}}{2} \)This simplifies to: \( r = \frac{4 \pm 4i}{2} \).So, \( r = 2 \pm 2i \).
04

Write the General Solution

The roots are complex conjugates \( 2 \pm 2i \). For complex roots in the form \( \alpha \pm \beta i \), the general solution is:\( y(x) = e^{\alpha x} (C_1 \cos{\beta x} + C_2 \sin{\beta x}) \).Here, \( \alpha = 2 \) and \( \beta = 2 \), so the general solution is:\( y(x) = e^{2x} (C_1 \cos{2x} + C_2 \sin{2x}) \).

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Key Concepts

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

Characteristic Equation
A second-order differential equation often appears in the form: \( a y^{\text{''}} + b y^{\text{'}} + cy = 0 \). This equation can be transformed into a characteristic equation to simplify the solving process.
The characteristic equation is a quadratic equation formed by replacing \( y^{\text{''}} \) with \( r^2 \), \( y^{\text{'}} \) with \( r \), and \( y \) with 1. Here鈥檚 how it's done:
  • Given the differential equation \( y^{\text{''}} - 4y^{\text{'}} + 8y = 0 \)
  • Transform it to \( r^2 - 4r + 8 = 0 \)
By solving this quadratic equation, you can find the values of \( r \) (called roots) that will help in constructing the general solution. This characteristic equation simplifies the differential equation to a solvable algebra problem.
Complex Roots
When solving the characteristic equation, you might encounter complex roots. Complex roots arise when the part under the square root in the quadratic formula is negative.
For the equation \( r^2 - 4r + 8 = 0 \):
  • Apply the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
  • In this scenario, it becomes \( r = \frac{4 \pm \sqrt{16 - 32}}{2} \)
  • This further simplifies to \( r = \frac{4 \pm \sqrt{-16}}{2} \)
  • The result is \( r = 2 \pm 2i \)
These complex roots are expressed in the form \( \alpha \pm \beta i \) where \( \alpha = 2 \) and \( \beta = 2 \). Knowing this form helps in the next step but ensure to remember that complex roots usually appear in conjugate pairs.
General Solution
After finding the roots of the characteristic equation, you can construct the general solution to the differential equation.
With complex roots \(\alpha \pm \beta i\), the solution to the differential equation takes a specific form:
The general solution is given by:
  • \( y(x) = e^{\alpha x} (C_1 \cos{\beta x} + C_2 \sin{\beta x}) \)
  • Here, \( \alpha = 2 \) and \( \beta = 2 \)
  • Plugging in these values, we get:
  • \( y(x) = e^{2x} (C_1 \cos{2x} + C_2 \sin{2x}) \)

This formula shows how the solution varies with \( x \). The constants \(C_1\) and \(C_2\) are determined by initial conditions or further context in specific problems. Understanding this helps you solve various second-order differential equations with complex roots.

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Most popular questions from this chapter

Tank A contains 100 gal of pure water. Tank B contains \(33 \mathrm{lb}\) of salt dissolved in 50 gal of water. Pure water is poured into tank \(B\) at the rate of \(3.5\) gal per min while an equal amount of the mixture is drained from the bottom of tank B. The mixture from tank \(A\) is pumped to tank \(B\) at the rate of 10 gal per min, while that from tank \(B\) is pumped to tank \(A\) at the same rate. Assume that the mixture in each tank is kept uniform by stirring. Let \(A(t)\) and \(B(t)\) be the amount of salt in tanks \(A\) and \(B\) after \(t\) minutes, respectively. a) Draw a two-compartment model for \(A(t)\) and \(B(t) .\) b) Show that \(A(t)\) and \(B(t)\) satisfy the differential equations \(A^{\prime}=-0.1 A+0.2 B\) and \(B^{\prime}=0.1 A-0.27 B\) c) Use the initial conditions \(A(0)=0\) and \(B(0)=33\) to solve for \(A\) and \(B\). d) Use a grapher to plot \(A(t)\) and \(B(t)\) for \(0 \leq t \leq 50\)

Suppose a reactant has two isomeric forms \(A\) and \(B\). The reactant initially has concentration \(q\) and is entirely in form \(A\), but may freely convert from form \(A\) to form \(B\) and back again. Also, once in form \(A\), the reactant may produce two products \(C\) and \(D\). The concentration \(F(t)\) of product \(C\) after \(t\) minutes satisfies the second-order differential equation $$ F^{\prime \prime}+(a+b+c+d) F^{\prime}+b(c+d) F=b c q $$ where the constants \(a, b, c\), and \(d\) describe the rates that \(A, B, C\), and \(D\) are produced. Also, \(F(0)=0\) and \(F^{\prime}(0)=c q\) initially. Suppose that \(a=0, b=1, c=0.05, d=0\), and \(q=1 .\) Find \(F(t)\)

Let \(x\) and \(y\) represent the populations (in thousands) of two species that share a habitat. For each system of equations: a) Find the equilibrium points and assess their stability. Solve only for equilibrium points representing nonnegative populations. b) Give the biological interpretation of the asymptotically stable equilibrium point(s). \(x^{\prime}=x(0.1-0.006 x-0.0008 y)\) \(y^{\prime}=y(0.2-0.001 x-0.006 y)\)

Find the general solution of the system of equations. \(x^{\prime}=y+2 t+3, y^{\prime}=-x+4 t-2\)

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