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Each equation has a characteristic equation possessing distinct real roots. Find the general solution of each equation. \(y^{(4)}+36 y=13 y^{\prime \prime}\)

Short Answer

Expert verified
The general solution is \( y(t) = C_1 e^{3t} + C_2 e^{-3t} + C_3 e^{2t} + C_4 e^{-2t} \).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is:\[y^{(4)} + 36y = 13y''\]Let's rewrite it as:\[y^{(4)} - 13y'' + 36y = 0\]
02

Assume a Solution Form

For linear differential equations with constant coefficients, assume a solution of the form:\[y = e^{rt}\]where \( r \) is a constant to be determined.
03

Find the Characteristic Equation

Substitute \( y = e^{rt} \) into the differential equation to obtain the characteristic equation:\[r^4 - 13r^2 + 36 = 0\]This can be rewritten as a quadratic in terms of \( r^2 \):\[x^2 - 13x + 36 = 0\]where \( x = r^2 \).
04

Solve the Quadratic Equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( x^2 - 13x + 36 = 0 \):\[x = \frac{13 \pm \sqrt{169 - 144}}{2}\]\[x = \frac{13 \pm 5}{2}\]This gives us solutions \( x = 9 \) and \( x = 4 \).
05

Determine the Values of r

Recall that \( x = r^2 \), substituting the values:- For \( x = 9 \), \( r^2 = 9 \), so \( r = \pm 3 \).- For \( x = 4 \), \( r^2 = 4 \), so \( r = \pm 2 \).
06

Write the General Solution

Since we have distinct real roots \( r = 3, -3, 2, -2 \), the general solution to the differential equation is:\[y(t) = C_1 e^{3t} + C_2 e^{-3t} + C_3 e^{2t} + C_4 e^{-2t}\]where \( C_1, C_2, C_3, \) and \( C_4 \) 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 key concept when solving linear differential equations with constant coefficients. It arises because of the method we use to determine the possible solutions of the differential equation. The idea is to convert the original differential equation into an algebraic equation, known as the characteristic equation, where the derivatives are replaced by powers of a variable, usually denoted as \( r \).

In the given differential equation:
  • We assumed a solution of the form \( y = e^{rt} \)
  • We substituted into the differential equation to get an equation in terms of \( r \)
  • This yielded the characteristic equation: \( r^4 - 13r^2 + 36 = 0 \)
This characteristic equation tells us the values of \( r \) for which our assumed solution form \( e^{rt} \) is a solution of the differential equation.
General Solution
The general solution of a differential equation is essential because it encompasses all possible solutions, given the linearity and rules of the differential equation. For linear differential equations with constant coefficients, once we have the characteristic equation:
  • We solve it to find the roots.
  • The nature of these roots (real and distinct, real and repeated, or complex) dictates the form of the general solution.
In this case, after solving the characteristic equation, we obtained distinct real roots \( r = 3, -3, 2, -2 \). These roots correspond to exponential solutions:\[y(t) = C_1 e^{3t} + C_2 e^{-3t} + C_3 e^{2t} + C_4 e^{-2t}\]where each \( C_i \) is an arbitrary constant. This formula represents the complete general solution for our differential equation.
Real Roots
Real roots in the context of the characteristic equation mean the solutions are exponential in nature. They tell us about the behavior and type of the function solutions. To determine these roots:
  • We solved the characteristic equation like a polynomial.
  • Substituted \( x = r^2 \), transforming the equation into a simple quadratic: \( x^2 - 13x + 36 = 0 \).
  • Using the quadratic formula, we found \( x = 9 \) and \( x = 4 \).
  • This gave us the roots \( r = \pm 3 \) and \( r = \pm 2 \), since \( r^2 = x \).
Distinct real roots lead to distinct terms in the general solution, providing a separate exponential term for each root.
Linear Differential Equations
Linear differential equations are equations involving a function and its derivatives with linear combinations. In such equations, each term is either a constant or the product of a constant and a function of the independent variable, without any multiplications between derivatives of the function.

The equation in the exercise, \( y^{(4)} + 36y = 13y'' \), is a linear differential equation. Here:
  • The highest order derivative is \( y^{(4)} \).
  • It includes simple linear terms of \( y \).
Because it is linear with constant coefficients, we could employ the technique of utilizing the characteristic equation to find the general solution efficiently. This property allows for systematic approaches to be applied, making it easier to solve compared to nonlinear differential equations.

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Most popular questions from this chapter

The matrix \(A\) has one real eigenvalue of multiplicity two. Find the general solution of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rr}-2 & 0 \\ 0 & -2\end{array}\right)\)

Let \(A\) be a \(2 \times 2\) matrix with a single eigenvalue \(\lambda\) of algebraic multiplicity 2 and geometric multiplicity 1. Prove that $$ e^{A t}=e^{\lambda t}[I+(A-\lambda I) t] . $$

Consider the system $$ \mathbf{y}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) \mathbf{y} . $$ (a) Find the eigenvalues and eigenvectors. Use your numerical solver to sketch the half-lines generated by the exponential solutions \(\mathbf{y}(t)=C_{i} e^{\lambda_{i} t} \mathbf{v}_{i}\) 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?

Suppose that \(A\) is a real \(2 \times 2\) matrix with one eigenvalue \(\lambda\) of multiplicity two. Show that the solution to the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}\) with \(\mathbf{y}(0)=\mathbf{v}\) is given by $$ \mathbf{y}(t)=e^{\lambda t}[\mathbf{v}+t(A-\lambda I) \mathbf{v}] . $$

Provides a general solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\), for some \(A\). Without the help of a computer or a calculator, sketch the half-line solutions generated by each exponential term of the solution. Then, sketch a rough approximation of a solution in each region determined by the half-line solutions. Use arrows to indicate the direction of motion on all solutions. Classify the equilibrium point as a saddle, a nodal sink, or a nodal source. \(\mathbf{y}(t)=C_{1} e^{t}\left(\begin{array}{l}-1 \\\ -2\end{array}\right)+C_{2} e^{2 t}\left(\begin{array}{r}3 \\\ -1\end{array}\right)\)

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