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Chapter 3: Problem 2

A homogeneous second-order linear differential equation, two functions \(y_{1}\) and \(y_{2}\), and a pair of initial conditions are given. First verify that \(y_{1}\) and \(y_{2}\) are solutions of the differential equation. Then find a particular solution of the form \(y=c_{1} y_{1}+c_{2} y_{2}\) that satisfies the given initial conditions. Primes denote derivatives with respect to \(x\). $$ y^{\prime \prime}-9 y=0 ; y_{1}=e^{3 x}, y_{2}=e^{-3 x} ; y(0)=-1, y^{\prime}(0)=15 $$

Short Answer

Expert verified
The particular solution is \( y = 2e^{3x} - 3e^{-3x} \).

Step by step solution

01

Verify that y1 and y2 are solutions

The differential equation is \( y'' - 9y = 0 \). For \( y_1 = e^{3x} \), calculate its derivatives: \( y_1' = 3e^{3x} \) and \( y_1'' = 9e^{3x} \). Substituting into the differential equation gives \( 9e^{3x} - 9e^{3x} = 0 \), verifying \( y_1 \) is a solution. Similarly, for \( y_2 = e^{-3x} \), calculate its derivatives: \( y_2' = -3e^{-3x} \) and \( y_2'' = 9e^{-3x} \). Substituting into the differential equation gives \( 9e^{-3x} - 9e^{-3x} = 0 \), verifying \( y_2 \) is also a solution.
02

Form the general solution

Since \( y_1 \) and \( y_2 \) are both solutions to the homogeneous differential equation, their linear combination \( y = c_1y_1 + c_2y_2 = c_1e^{3x} + c_2e^{-3x} \) is the general solution of the differential equation.
03

Apply the initial condition y(0) = -1

Substitute \( x = 0 \) into the general solution: \( y(0) = c_1e^{0} + c_2e^{0} = c_1 + c_2 = -1 \). This gives us the first condition: \( c_1 + c_2 = -1 \).
04

Apply the initial condition y'(0) = 15

First, determine the expression for the derivative: \( y' = 3c_1e^{3x} - 3c_2e^{-3x} \). Substitute \( x = 0 \) into this equation: \( y'(0) = 3c_1 - 3c_2 = 15 \). This gives the second condition: \( 3c_1 - 3c_2 = 15 \).
05

Solve the system of equations

You now have two equations: \( c_1 + c_2 = -1 \) and \( 3c_1 - 3c_2 = 15 \). Solve the second equation for \( c_1 \) by dividing by 3, yielding \( c_1 - c_2 = 5 \). Now solve the system of equations: 1) \( c_1 + c_2 = -1 \) 2) \( c_1 - c_2 = 5 \). Adding these equations, \( 2c_1 = 4 \) which gives \( c_1 = 2 \). Substituting \( c_1 = 2 \) into \( c_1 + c_2 = -1 \) gives \( 2 + c_2 = -1 \), hence \( c_2 = -3 \).
06

Construct the particular solution

Substitute \( c_1 = 2 \) and \( c_2 = -3 \) back into the general solution to obtain the particular solution: \( y = 2e^{3x} - 3e^{-3x} \).

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Key Concepts

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

Homogeneous Differential Equations
Homogeneous differential equations are equations where every term is a multiple of the function or its derivatives, and there's no term that stands alone (such as a constant term). The concept here is to find solutions where the function itself doesn't force the equation to fail unless the expressions involving the function can totally sum to zero on their own.

In the case of second-order linear differential equations like this one, the simplicity of the structure permits solutions of exponential forms, such as the ones given here:
  • The differential equation given is \( y'' - 9y = 0 \).
  • For the solutions \( y_1 = e^{3x} \) and \( y_2 = e^{-3x} \), each term in the equation contains the function \( y \) or its derivatives. Due to this property, these solutions will naturally satisfy the neutralization needed for a homogeneous equation.
  • Showing that these trial solutions nullify the differential equation is essentially what verifying solutions means in this context.
Initial Value Problem
An initial value problem involves finding a specific solution to a differential equation that satisfies initial conditions, providing specific values for the solution and possibly its derivatives at a particular point, usually \( x = 0 \).

In our exercise, two initial conditions are presented:
  • The solution must satisfy \( y(0) = -1 \), meaning when \( x = 0 \), the function's output should be \(-1\).
  • Additionally, the first derivative \( y'(0) = 15 \) must hold, prescribing that the rate of change of the function at zero should equal 15.
These initial conditions form a system of equations to determine the constants in the particular solution, guiding the solution to meet these constraints explicitly.
Particular Solution
A particular solution is a specific solution to the differential equation which meets the given initial conditions.
  • A general solution to a homogeneous differential equation usually involves arbitrary constants, which allow for a family of solutions including the true particular ones fitting specific conditions.
  • In this case, the general solution is \( y = c_1 e^{3x} + c_2 e^{-3x} \).
To pinpoint the particular solution,
  • Apply each of the initial conditions to establish values for the constants \( c_1 \) and \( c_2 \).
  • Solving the set of equations from the initial conditions reveals \( c_1 = 2 \) and \( c_2 = -3 \), leading us to the particular solution: \( y = 2e^{3x} - 3e^{-3x} \).
Verification of Solutions
Verification involves confirming that chosen solutions indeed nullify the given differential equation, ensuring both trial solutions and other expressions satisfy the initial problem setup.
  • For any proposed solution like \( y_1 \) or \( y_2 \), verify by substitution back into the original differential equation and check that the equation holds.
  • For example, substitute \( y_1 = e^{3x} \) back through to compute its derivatives and substitute again inside the original equation \( y'' - 9y = 0 \).This must sum to zero to confirm \( y_1 \) is a valid solution.
  • Similarly, ensure your particular solution \( y = 2e^{3x} - 3e^{-3x} \) adheres to the original differential equation and meets any and all given initial conditions effectively.
Verification is about being confident that the mathematics functions correctly and predictably in models, whether abstract or real-world applications.

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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. $$ y^{(4)}+y^{(3)}-3 y^{\prime \prime}-5 y^{\prime}-2 y=0 $$

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 \() .\) $$ m=1, c=4, k=5, F_{0}=10 $$

Assume as known that the Vandermonde determinant $$ V=\left|\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ r_{1} & r_{2} & \cdots & r_{n} \\ r_{1}^{2} & r_{2}^{2} & \cdots & r_{n}^{2} \\ \vdots & \vdots & & \vdots \\ r_{1}^{n-1} & r_{2}^{n-1} & \cdots & r_{n}^{n-1} \end{array}\right| $$ is nonzero if the numbers \(r_{1}, r_{2}, \ldots, r_{n}\) are distinct. Prove by the method of Problem 33 that the functions $$ f_{i}(x)=\exp \left(r_{i} x\right), \quad 1 \leqq i \leqq n $$ are linearly independent.

Find the steady periodic solution \(x_{\mathrm{sp}}(t)=C \cos (\omega t-\alpha)\) of the given equation \(m x^{\prime \prime}+\) \(c x^{\prime}+k x=F(t)\) with periodic forcing function \(F(t)\) of frequency \(\omega .\) Then graph \(x_{\mathrm{sp}}(t)\) together with (for comparison) the adjusted forcing function \(F_{1}(t)=F(t) / m \omega .\) $$ x^{\prime \prime}+4 x^{\prime}+4 x=10 \cos 3 t $$

Show that \(y_{1}=\sin x^{2}\) and \(y_{2}=\cos x^{2}\) are linearly independent functions, but that their Wronskian vanishes at \(x=0\). Why does this imply that there is no differential equation of the form \(y^{\prime \prime}+p(x) y^{\prime}+q(x) y=0\), with both \(p\) and \(q\) continuous everywhere, having both \(y_{1}\) and \(y_{2}\) as solutions?
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