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Chapter 3: Problem 48

In Problems 47 through 56, use the method of variation of parameters to find a particular solution of the given differential equation. $$ y^{\prime \prime}-2 y^{\prime}-8 y=3 e^{-2 x} $$

Short Answer

Expert verified
The particular solution is \( y_p(x) = \frac{1}{2}x e^{4x} - \frac{1}{4} \) and the general solution is \( y(x) = C_1 e^{4x} + C_2 e^{-2x} + \frac{1}{2}x e^{4x} - \frac{1}{4} \).

Step by step solution

01

Find the Homogeneous Solution

Solve the homogeneous version of the equation: \[ y'' - 2y' - 8y = 0 \]The characteristic equation is given by:\[ r^2 - 2r - 8 = 0 \]Factor the quadratic to obtain:\[ (r - 4)(r + 2) = 0 \]The roots are \( r = 4 \) and \( r = -2 \). Thus, the general solution of the homogeneous equation is:\[ y_h(x) = C_1 e^{4x} + C_2 e^{-2x} \]
02

Determine Functions for Variation of Parameters

For the method of variation of parameters, find two solutions from the homogeneous solution:\[ y_1(x) = e^{4x}, \ y_2(x) = e^{-2x} \]
03

Compute the Wronskian

The Wronskian \( W(y_1, y_2) \) is computed as:\[ W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]Substitute the functions and their derivatives:\[ W = \begin{vmatrix} e^{4x} & e^{-2x} \ 4e^{4x} & -2e^{-2x} \end{vmatrix} \]Simplifying, we find:\[ W = (e^{4x})(-2e^{-2x}) - (4e^{4x})(e^{-2x}) \]\[ W = -2e^{2x} - 4e^{2x} = -6e^{2x} \]
04

Integrate to Find Particular Solution Coefficients

Using the method of variation of parameters, determine the integrals:\[ v_1'(x) = \frac{-y_2(x)g(x)}{W} = \frac{-e^{-2x} \cdot 3e^{-2x}}{-6e^{2x}} \]Simplifying, we get:\[ v_1'(x) = \frac{3}{6} = \frac{1}{2} \]\[ v_1(x) = \int \frac{1}{2} \, dx = \frac{1}{2}x + C_1 \]Similarly, find \( v_2'(x) \):\[ v_2'(x) = \frac{y_1(x)g(x)}{W} = \frac{e^{4x} \cdot 3e^{-2x}}{-6e^{2x}} \]\[ v_2'(x) = -\frac{1}{2}e^{2x} \]\[ v_2(x) = -\frac{1}{2} \int e^{2x} \, dx = -\frac{1}{4}e^{2x} + C_2 \]
05

Assemble the Particular Solution

The particular solution \( y_p(x) \) is given by:\[ y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x) \]Substitute \( v_1(x) \) and \( v_2(x) \):\[ y_p(x) = \left(\frac{1}{2}x \right)e^{4x} + \left(-\frac{1}{4}e^{2x}\right)e^{-2x} \]Simplify:\[ y_p(x) = \frac{1}{2}x e^{4x} - \frac{1}{4} \]
06

Write the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:\[ y(x) = y_h(x) + y_p(x) \]\[ y(x) = C_1 e^{4x} + C_2 e^{-2x} + \frac{1}{2}x e^{4x} - \frac{1}{4} \]

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve the derivatives of a function. These equations are fundamental in describing various physical phenomena such as heat conduction, wave propagation, and population dynamics. A typical differential equation is expressed as a relation involving the function and its derivatives.

In the context of our exercise, the differential equation we are dealing with is a second-order linear differential equation:
  • Second-order: It involves up to the second derivative of the function, denoted as \( y'' \).
  • Linear: The function and its derivatives appear to the power of one, and are not multiplied together.
This type of equation is solved by finding functions that satisfy the relation for all values in a certain interval, allowing us to predict the behavior of the system modeled by the equation.
Particular Solution
The particular solution, denoted as \( y_p(x) \), is essential in forming the complete solution of non-homogeneous differential equations like the one in our exercise.

The non-homogeneous differential equation has the form \( y'' - 2y' - 8y = 3e^{-2x} \), where \( 3e^{-2x} \) is a function causing the equation to be non-homogeneous.

To find the particular solution, we employ the variation of parameters method, which involves:
  • Identifying and integrating specific functions derived from the homogeneous solution.
  • Substituting these integrated functions back into a particular form to match the non-homogeneous equation.
By constructing \( y_p(x) \) using the solutions \( v_1(x) \) and \( v_2(x) \), we address the specific 'forcing' added by the non-homogeneous term.
Wronskian
The Wronskian is a determinant useful for verifying whether a set of solutions is linearly independent. It is fundamental in applying the variation of parameters.

Given two functions \( y_1(x) \) and \( y_2(x) \), the Wronskian \( W \) is formulated as:\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix}\]For our exercise, substituting \( y_1(x) = e^{4x} \) and \( y_2(x) = e^{-2x} \):
  • Calculate derivatives: \( y_1' = 4e^{4x} \) and \( y_2' = -2e^{-2x} \).
  • Substitute into the determinant and simplify.
The result, \( W = -6e^{2x} \), indicates the functions are linearly independent and can be used in forming the solution.
Homogeneous Solution
Generating a homogeneous solution is a critical step in solving differential equations. For our equation, \( y'' - 2y' - 8y = 0 \) describes the homogeneous part, where the equation equals zero.

The homogeneous solution, denoted \( y_h(x) \), is derived by solving an associated characteristic equation:
  • Set up the characteristic polynomial: \( r^2 - 2r - 8 = 0 \).
  • Factor to \( (r-4)(r+2) = 0 \), yielding roots \( r = 4 \) and \( r = -2 \).
The general homogeneous solution becomes \( y_h(x) = C_1 e^{4x} + C_2 e^{-2x} \). This solution forms the basis for superimposing particular solutions to account for the non-homogeneous part of the equation.

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Most popular questions from this chapter

Find general solutions of the equations in Problems. First find a small integral root of the characteristic equation by inspection; then factor by division. $$ y^{(3)}+3 y^{\prime \prime}+4 y^{\prime}-8 y=0 $$

Suppose that the three numbers \(r_{1}, r_{2}\), and \(r_{3}\) are distinct. Show that the three functions \(\exp \left(r_{1} x\right), \exp \left(r_{2} x\right)\), and \(\exp \left(r_{3} x\right)\) are linearly independent by showing that their Wronskian $$ W=\exp \left[\left(r_{1}+r_{2}+r_{3}\right) x\right] \cdot\left|\begin{array}{ccc} 1 & 1 & 1 \\ r_{1} & r_{2} & r_{3} \\ r_{1}^{2} & r_{2}^{2} & r_{3}^{2} \end{array}\right| $$ is nonzero for all \(x\).

As indicated by the cart-with-flywheel example discussed in this section, an unbalanced rotating machine part typically results in a force having amplitude proportional to the square of the frequency \(\omega\). (a) Show that the amplitude of the steady periodic solution of the differential equation $$ m x^{\prime \prime}+c x^{\prime}+k x=m A \omega^{2} \cos \omega t $$ (with a forcing term similar to that in Eq. (17)) is given by $$ C(\omega)=\frac{m A \omega^{2}}{\sqrt{\left(k-m \omega^{2}\right)^{2}+(c \omega)^{2}}}. $$ (b) Suppose that \(c^{2}<2 m k .\) Show that the maximum amplitude occurs at the frequency \(\omega_{m}\) given by $$ \omega_{m}=\sqrt{\frac{k}{m}\left(\frac{2 m k}{2 m k-c^{2}}\right)} . $$ Thus the resonance frequency in this case is larger (in contrast with the result of Problem 27 ) than the natural frequency \(\omega_{0}=\sqrt{k / m} .\) (Suggestion: Maximize the square of \(C .\) )

Gives the parameters for a forced mass-spring-dashpot system with equation \(m x^{\prime \prime}+\) \(c x^{\prime}+k x=F_{0} \cos \omega t .\) Investigate the possibility of practical resonance of this system. In particular, find the amplitude \(C(\omega)\) of steady periodic forced oscillations with frequency \(\omega .\) Sketch the graph of \(C(\omega)\) and find the practical resonance frequency \(\omega(\) if any \() .\) $$ m=1, c=10, k=650, F_{0}=100 $$

In Problems 21 through 24, a nonhomogeneous differential equation, a complementary solution \(y_{c}\), and a particular solution \(y_{p}\) are given. Find a solution satisfying the given initial conditions. $$ \begin{aligned} &y^{\prime \prime}-4 y=12 ; y(0)=0, y^{\prime}(0)=10 ; \\ &y_{c}=c_{1} e^{2 x}+c_{2} e^{-2 x} ; y_{p}=-3 \end{aligned} $$
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