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Chapter 6: Problem 5

Find the general solution of each of the following systems: (a) \(y_{1}^{\prime \prime}=-2 y_{2}\) (b) \(y_{1}^{\prime \prime}=2 y_{1}+y_{2}^{\prime}\) \(y_{2}^{\prime \prime}=y_{1}+3 y_{2} \quad y_{2}^{\prime \prime}=2 y_{2}+y_{1}^{\prime}\)

Short Answer

Expert verified
The general solutions for the given systems are: (a) \(y_1(t) = At + B\), \(y_2(t) = 0\). (b) \(y_1(t) = C_1 \cos (\sqrt{2}t) + C_2 \sin (\sqrt{2}t) + \frac{1}{3}t\), \(y_2(t) = C_3 \cos (\sqrt{5}t) + C_4 \sin (\sqrt{5}t) - \frac{1}{6}t\), where A, B, \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are arbitrary constants.

Step by step solution

01

Rewrite the equation

Rewrite the given equation as follows: \(y_1^{\prime \prime} + 2y_2 = 0\). Step 2: Solve the homogeneous equation for \(y_1\)
02

Solve the homogeneous equation

To find the general solution of the homogeneous equation \(y_1^{\prime \prime} = 0\), we must find a function whose second derivative is zero. This will be a linear function of the form \(y_1(t) = At + B\), where A and B are arbitrary constants. Step 3: Differentiate \(y_1(t)\)
03

Differentiate the general solution

In order for us to find the relationship between \(y_1\) and \(y_2\), we must differentiate \(y_1(t)\) to get \(y_1^{\prime}(t) = A\). Step 4: Find \(y_2(t)\)
04

Find the general solution for \(y_2\)

Plug \(y_1^{\prime \prime}(t) = 0\) back into the original equation \(y_1^{\prime \prime} + 2y_2 = 0\) to find \(y_2(t)\). Therefore, we have \(2y_2 = 0\), which implies \(y_2(t) = 0\). The general solution for the given system (a) is \(y_1(t) = At + B\) and \(y_2(t) = 0\). #Exercise (b)# Step 1: Rewrite the given equations
05

Rewrite the equations

Rewrite the given equations as follows: 1. \(y_1^{\prime \prime} - 2y_1 - y_2^{\prime} = 0\) 2. \(y_2^{\prime \prime} - y_1 - 3y_2 = 0\) 3. \(y_2^{\prime \prime} - 2y_2 - y_1^{\prime} = 0\) Step 2: Eliminate \(y_1^{\prime}\) and \(y_2^{\prime}\) from the equations
06

Elimination phase

Subtract equation 3 from equation 1 to eliminate \(y_1^{\prime}\) and subtract equation 2 from equation 3 to eliminate \(y_2^{\prime}\): 4. \(2y_1^{\prime \prime} + y_1 - 2y_2 = 0\) 5. \(y_1^{\prime \prime} + y_1 - 5y_2 = 0\) Step 3: Solve the resulting equations
07

Solve for \(y_1(t)\) and \(y_2(t)\)

We can now solve equation 4 for \(y_1\) and equation 5 for \(y_2\). The general solutions for these differential equations are: \(y_1(t) = C_1 \cos (\sqrt{2}t) + C_2 \sin (\sqrt{2}t) + \frac{1}{3}t\) \(y_2(t) = C_3 \cos (\sqrt{5}t) + C_4 \sin (\sqrt{5}t) - \frac{1}{6}t\) Step 4: Combine the solutions
08

Combine the solutions

The general solution for the given system (b) is: \(y_1(t) = C_1 \cos (\sqrt{2}t) + C_2 \sin (\sqrt{2}t) + \frac{1}{3}t\) \(y_2(t) = C_3 \cos (\sqrt{5}t) + C_4 \sin (\sqrt{5}t) - \frac{1}{6}t\) Where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are arbitrary constants.

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Most popular questions from this chapter

Show that if \(A\) is a symmetric positive definite matrix, then \(A\) is nonsingular and \(A^{-1}\) is also positive definite.

Let \(B\) be an \(m \times n\) matrix of rank \(n .\) Show that \(B^{T} B\) is positive definite.

Let \\[ A=\left(\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{5} \\ \frac{1}{4} & \frac{1}{3} & \frac{2}{5} \\ \frac{1}{4} & \frac{1}{3} & \frac{2}{5} \end{array}\right) \\] be a transition matrix for a Markov process. (a) Compute det \((A)\) and trace \((A)\) and make use of those values to determine the eigenvalues of \(A\) (b) Explain why the Markov process must converge to a steady-state vector. (c) Show that \(\mathbf{y}=(16,15,15)^{T}\) is an eigenvector of \(A .\) How is the steady-state vector related to \(\mathbf{y} ?\)

Let \(A\) be a \(n \times n\) matrix with real entries and let \(\lambda_{1}=a+b i\) (where \(a\) and \(b\) are real and \(b \neq 0\) ) be an eigenvalue of \(A .\) Let \(\mathbf{z}_{1}=\mathbf{x}+i \mathbf{y}\) (where \(\mathbf{x}\) and \(\mathbf{y}\) both have real entries) be an eigenvector belonging to \(\lambda_{1}\) and let \(\mathbf{z}_{2}=\mathbf{x}-i \mathbf{y}\) (a) Explain why \(\mathbf{z}_{1}\) and \(\mathbf{z}_{2}\) must be linearly independent. (b) Show that \(\mathbf{y} \neq \mathbf{0}\) and that \(\mathbf{x}\) and \(\mathbf{y}\) are linearly independent.

Let \(\lambda\) be an eigenvalue of an \(n \times n\) matrix \(A\) and let \(\mathbf{x}\) be an eigenvector belonging to \(\lambda .\) Show that \(e^{\lambda}\) is an eigenvalue of \(e^{A}\) and \(\mathbf{x}\) is an eigenvector of \(e^{A}\) belonging to \(e^{\lambda}\)
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