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Each equation has a characteristic equation possessing distinct real roots. Find the general solution of each equation. \(y^{\prime \prime \prime}+30 y=4 y^{\prime \prime}+11 y^{\prime}\)

Short Answer

Expert verified
The general solution is \(y(x) = C_1 e^{x} + C_2 e^{3x} + C_3 e^{-10x}\).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \(y^{\prime \prime \prime} + 30y = 4y^{\prime \prime} + 11y^{\prime}\). First, rearrange terms to form a standard differential equation: \(y^{\prime \prime \prime} - 4y^{\prime \prime} - 11y^{\prime} + 30y = 0\).
02

Write the Characteristic Equation

For a third-order differential equation like \(y^{\prime \prime \prime} - 4y^{\prime \prime} - 11y^{\prime} + 30y = 0\), substitute \(y = e^{rx}\) to form the characteristic equation: \(r^3 - 4r^2 - 11r + 30 = 0\).
03

Find the Roots of the Characteristic Equation

Solve the characteristic equation \(r^3 - 4r^2 - 11r + 30 = 0\) for its roots. Use the Rational Root Theorem or synthetic division to find that its roots are \(r = 1, r = 3, r = -10\).
04

Write the General Solution

Since the roots of the characteristic equation are distinct and real, the general solution of the differential equation can be written as \(y(x) = C_1 e^{x} + C_2 e^{3x} + C_3 e^{-10x}\), where \(C_1, C_2,\) and \(C_3\) are arbitrary constants.

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Key Concepts

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

Characteristic Equation
The characteristic equation is a fundamental tool used in solving linear differential equations. In the case of a third-order differential equation like the one given: \[ y^{\prime \prime \prime} - 4y^{\prime \prime} - 11y^{\prime} + 30y = 0, \]we need to rewrite it in terms of an exponential function substitution. By assuming a solution of the form \( y = e^{rx} \), we can replace the derivatives in the equation by powers of \( r \), ultimately arriving at the characteristic polynomial:\[ r^3 - 4r^2 - 11r + 30 = 0. \]This polynomial equation is called the characteristic equation. Solving it gives us the values of \( r \) that characterize the behavior of solutions to the differential equation.
Distinct Real Roots
When solving the characteristic equation \( r^3 - 4r^2 - 11r + 30 = 0 \), we found its roots to be \( r = 1, r = 3, \) and \( r = -10 \). These roots are distinct, meaning they are unique and do not repeat. Having distinct real roots provides us with a straightforward way to construct the general solution. Each distinct root corresponds to an exponential component in the solution of the differential equation. This results in a solution that is a simple linear combination of these exponential terms.
General Solution
With the distinct real roots \( r = 1, r = 3, \) and \( r = -10 \), the general solution to the differential equation can be expressed as:\[ y(x) = C_1 e^x + C_2 e^{3x} + C_3 e^{-10x}, \]where \( C_1, C_2, \) and \( C_3 \) are arbitrary constants. These constants will be determined based on initial conditions or boundary values given in specific problems. Each term in this solution corresponds directly to one of the roots of the characteristic equation, highlighting how the distinct and real nature of the roots simplifies the form of the solution.
Rational Root Theorem
The Rational Root Theorem is a handy technique for finding the possible rational roots of a polynomial equation. It was crucial in solving our problem's characteristic equation \( r^3 - 4r^2 - 11r + 30 = 0 \). This theorem states that any rational root of a polynomial, with integer coefficients, is in the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. In our example, the constant term is 30, and the leading coefficient is 1.
  • Factors of 30: \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 \)
  • Factors of 1: \( \pm 1 \)
Using this theorem provided us a manageable list of potential rational roots that we could test, leading us efficiently to the roots \( r = 1, r = 3, \) and \( r = -10 \). Once these roots were identified, we could apply them directly in deriving the general solution.

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Most popular questions from this chapter

Consider the system $$ \begin{aligned} &x^{\prime}=-3 x \\ &y^{\prime}=-2 x-y \\ &z^{\prime}=-2 z \end{aligned} $$ (a) Find the eigenvalues and eigenvectors. Use your numerical solver to sketch the half-lines generated by the exponential solutions \(\mathbf{y}(t)=C_{i} e^{\lambda_{i} /} \mathbf{v}_{i}\) for \(i=1,2\), and \(3 .\) (b) The six half-line solutions found in part (a) come in pairs that form straight lines. These three lines, taken two at a time, generate three planes. In turn these planes divide phase space into eight octants. Use your numerical solver to add solution trajectories with initial conditions in each of the eight octants. (c) Based on your phase portrait, what would be an appropriate name for the equilibrium point in this case?

Find a fundamental set of solutions for the given system. Can be done by hand, but use a computer for the rest. \(\mathbf{x}^{\prime}=\left(\begin{array}{rrrrr}-2 & 2 & -2 & 0 & -3 \\ -1 & 0 & -1 & 0 & -3 \\ 15 & -16 & -1 & 10 & 33 \\ 12 & -13 & 1 & 6 & 26 \\ -5 & 5 & 0 & -3 & -12\end{array}\right) \mathbf{x}\)

Do the following for each of the matrices in Exercises 26-33. Exercises \(26-29\) can be done by hand, but you should use a computer for the rest. (i) Find the eigenvalues. (ii) For each eigenvalue, find the algebraic and the geometric multiplicities. (iii) For each eigenvalue \(\lambda\), find the smallest integer \(k\) such that the dimension of the nullspace of \((A-\lambda I)^{k}\) is equal to the algebraic multiplicity. (iv) For each eigenvalue \(\lambda\), find \(q\) linearly independent generalized eigenvectors, where \(q\) is the algebraic multiplicity of \(\lambda\). (v) Verify that the collection of the generalized eigenvectors you find in part (iv) for all of the eigenvalues is linearly independent. (vi) Find a fundamental set of solutions for the system \(\mathbf{y}^{\prime}=\) Ay. \(A=\left(\begin{array}{rrrr}18 & -7 & 24 & 24 \\ 15 & -8 & 20 & 16 \\ 0 & 0 & -1 & 0 \\ -12 & 4 & -15 & -17\end{array}\right)\)

For the \(2 \times 2\) matrices, use \(p(\lambda)=\) \(\lambda^{2}-T \lambda+D\), where \(T=\operatorname{tr}(A)\) and \(D=\operatorname{det}(A)\), to compute the characteristic polynomial. Then, use \(p(\lambda)=\operatorname{det}(A-\lambda I)\) to calculate the characteristic polynomial a second time and compare the results. \(A=\left(\begin{array}{rr}0 & 5 \\ -1 & 4\end{array}\right)\)

The matrix \(A\) has one real eigenvalue of multiplicity two. Find the general solution of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rr}3 & -1 \\ 1 & 1\end{array}\right)\)

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