/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 \(\mathbf{x}^{\prime}=\left[ \be... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 35

\(\mathbf{x}^{\prime}=\left[ \begin{array}{lll}{0} & {0} & {0} \\ {0} & {1} & {0} \\ {1} & {0} & {1}\end{array}\right] \mathbf{x}, \quad \mathbf{x}(t)=\left[ \begin{array}{c}{0} \\ {e^{t}} \\ {-3 e^{t}}\end{array}\right]\)

Short Answer

Expert verified
The given vector function \(x(t) = [0, e^t, -3e^t]^\intercal\) is indeed a solution to the given linear system of differential equations \(x' = Mx\) where \(M = [[0,0,0],[0,1,0],[1,0,1]]\) as its derivative is identical to the result of the matrix-vector multiplication.

Step by step solution

01

Compute the Derivative of \(x(t)\)

First, differentiate each component of the vector function \(x(t)\) with respect to \(t\). The result is \(x'(t) = [0, e^t, -3e^t]^\intercal\).
02

Compute Matrix-Vector multiplication \(Mx\)

Using the given matrix \(M\) and vector \(x(t)\), perform a matrix-vector multiplication. The result is \(Mx = [0, e^t, -3e^t]^\intercal\).
03

Compare \(x'(t)\) and \(Mx\)

Compare the derivative of \(x(t)\) from Step 1 and the product \(Mx\) from Step 2. Since they are identical, this verifies that \(x(t)\) is indeed a solution to the differential equation \(x' = Mx\).

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Most popular questions from this chapter

Illustrate the equivalence of the assertions (a)-(d) in Theorem 1 (page 511\()\) for the matrix \(\left[ \begin{array}{ccc}{4} & {-2} & {2} \\ {-2} & {4} & {2} \\\ {2} & {2} & {4}\end{array}\right]\) as follows. (a) Show that the row-reduction procedure applied to \([\mathbf{A} : \mathbf{I}]\) fails to produce the inverse of \(\mathbf{A} .\) (b) Calculate det \(\mathbf{A} .\) (c) Determine a nontrivial solution \(\mathbf{x}\) to \(\mathbf{A} \mathbf{x}=\mathbf{0}\) . (d) Find scalars \(c_{1}, c_{2},\) and \(c_{3},\) not all zero, so that \(c_{1} \mathbf{a}_{1}+c_{2} \mathbf{a}_{2}+c_{3} \mathbf{a}_{3}=\mathbf{0},\) where \(\mathbf{a}_{1}, \mathbf{a}_{2},\) and \(\mathbf{a}_{3}\) are the columns of \(\mathbf{A} .\)

\(\mathbf{x}^{\prime}(t)=\left[ \begin{array}{rr}{6} & {-3} \\ {2} & {1}\end{array}\right] \mathbf{x}(t), \quad \mathbf{x}(0)=\left[ \begin{array}{r}{-10} \\ {-6}\end{array}\right]\)

$$ \mathbf{x}^{\prime}(t)=\left[ \begin{array}{rr}{-1} & {-1} \\ {9} & {-1}\end{array}\right] \mathbf{x}(t) $$

\(\mathbf{A}=\left[ \begin{array}{ll}{1} & {1} \\ {4} & {1}\end{array}\right]\)

34\. Prove that a fundamental solution set for the homoge- neous system $$\mathbf{x}^{\prime}(t)=\mathbf{A}(t) \mathbf{x}(t)$$ always exists on an interval \(I,\) provided \(\mathbf{A}(t)\) is continuous on \(I .\) [Hint: Use the existence and uniqueness theorem (Theorem 2\()\) and make judicious choices for \(\mathbf{x}_{0} . ]\)
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