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Chapter 7: Problem 14

Verify that the given vector is the general solution of the corresponding homogeneous system, and then solve the non-homogeneous system. Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{1-t^{2}} \\ {2 t}\end{array}\right), \quad \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1} $$

Short Answer

Expert verified
Question: Find the complete solution of the non-homogeneous system given the general solution of the corresponding homogeneous system and the non-homogeneous equation: Homogeneous equation: $$ t\mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \mathbf{x} $$ General solution of the corresponding homogeneous system: $$ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1} $$ Non-homogeneous equation: $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{1-t^{2}} \\\ {2 t}\end{array}\right) $$ Answer: The complete solution of the non-homogeneous system is: $$ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1}+\frac{3t^{-1}(4t^{2}+3t+2)}{4}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+\frac{-3(4t^{2}+2t+1)}{4t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) $$

Step by step solution

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01

Verify the given vector is the general solution of the homogeneous system.

To verify that the given vector is the general solution of the corresponding homogeneous system, we will plug it into the homogeneous equation and check if it satisfies the equation. The given homogeneous equation is: $$ t\mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \mathbf{x} $$ and the given general solution is: $$ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1} $$ Let's find the derivative of \(\mathbf{x}^{(t)}\) with respect to t: $$ \mathbf{x}^{(t)\prime}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) (-t^{-2}) $$ Now we can plug the given vector and its derivative into the homogeneous equation: $$ t\cdot\left(c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) (-t^{-2})\right) = \left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \cdot \left(c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1}\right) $$ Simplifying, we find that both sides of the equation are equal, verifying that the given vector is the general solution of the corresponding homogeneous system.
02

Solve the non-homogeneous system

To solve the non-homogeneous system, we will substitute the given general solution into the non-homogeneous equation and find the particular solution. The given non-homogeneous equation is: $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{1-t^{2}} \\\ {2 t}\end{array}\right) $$ Substituting the given general solution and its derivative, we get: $$ t\cdot \left(c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) (-t^{-2})\right)=\left(\begin{array}{cc}{2} & {-1} \\\ {3} & {-2}\end{array}\right) \cdot \left(c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1}\right)+\left(\begin{array}{c}{1-t^{2}} \\\ {2 t}\end{array}\right) $$ Solving this system of equations for c1 and c2, we find that: $$ c_{1}=\frac{3t^{-1}(4t^{2}+3t+2)}{4}, \quad c_{2}=\frac{-3(4t^{2}+2t+1)}{4t} $$ Now we have the particular solution in the form of a first order linear system with undetermined coefficients: $$ \mathbf{x}^{(t)}=\frac{3t^{-1}(4t^{2}+3t+2)}{4}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+\frac{-3(4t^{2}+2t+1)}{4t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) $$ Combining the particular solution with the earlier found homogeneous general solution, we obtain the complete solution of the non-homogeneous system: $$ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1}+\frac{3t^{-1}(4t^{2}+3t+2)}{4}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+\frac{-3(4t^{2}+2t+1)}{4t}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) $$

Key Concepts

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

general solution
A general solution to a differential equation includes the complete set of possible solutions. In the context of a linear homogeneous system, this is represented by the linear combination of solutions. Consider a differential equation of the form: \[ t \mathbf{x}^{\prime} = \left(\begin{array}{cc}{2} & {-1} \ {3} &{-2}\end{array}\right) \mathbf{x} \] The general solution is expressed as: \[ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1}\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \ {3}\end{array}\right) t^{-1} \] This solution results from solving the system's characteristic equation and finding the eigenvectors associated with any eigenvalues. The constants \(c_1\) and \(c_2\) are determined by initial conditions. In real-world problems, the general solution offers flexibility, containing all potential solutions until specific conditions are applied.
homogeneous system
A homogeneous system of differential equations is characterized by having its non-homogeneous part (also known as the forcing term) equal to zero. It appears in the form: \[ t \mathbf{x}^{\prime} = A \mathbf{x} \] where \(A\) is the coefficient matrix. Such systems are important because they form the basis for understanding non-homogeneous systems. By solving homogeneous systems, we can determine the complimentary solution, which is part of the more complex solution of a non-homogeneous system.
  • Homogeneous systems are easier to solve because they only involve eigenvalues and eigenvectors.
  • The solution will typically result in exponential functions or linear combinations involving exponential functions.
This basic understanding transfers over when addressing the non-homogeneous equation. Essentially, the homogeneous solution serves as the foundation, adding particular solutions to account for the non-homogeneous parts.
first order linear system
A first-order linear system involves equations where the highest derivative is first-order. Mathematically, these systems involve equations of the form: \[ t \mathbf{x}^{\prime} = A \mathbf{x} + \mathbf{b}(t) \] where \(A\) is a matrix of coefficients and \(\mathbf{b}(t)\) is a function of \(t\), representing the non-homogeneous part. These systems are predominant in engineering and physics due to their real-world relevance.
  • These equations predict how systems evolute over time.
  • Linear systems can often be solved analytically, helping predict future behavior.
In first-order systems, it's integral to approach the solution methodically, starting with solving the homogeneous part and subsequently finding a particular solution that satisfies the entire system. Then, the overall solution becomes a combination of these.
vector solutions
Vector solutions are solutions expressed in vector form, which captures multiple variables simultaneously. This approach is especially useful for systems of differential equations that involve more than one interdependent quantity evolving over time. Consider the vector representation of our previously defined problem: \[ \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1}\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \ {3}\end{array}\right) t^{-1} \] Each part of the vector (in this case, \(\begin{array}{c}{1}\ {1}\end{array}\) and \(\begin{array}{c}{1} \ {3}\end{array}\)) shows how system variables correlate and evolve.
  • Vectors make it intuitive to see relationships between variables.
  • Efficient computational methods exist to handle vector forms, benefiting large systems.
Thus, approaching by vector solutions is not just advantageous for understanding individual-variable dynamics, but also in computing the solutions quickly and accurately using software tools.

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Most popular questions from this chapter

Solve the given system of equations by the method of Problem 19 of Section \(7.5 .\) Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$

In this problem we show that the eigenvalues of a Hermitian matrix \(\Lambda\) are real. Let \(x\) be an eigenvector corresponding to the eigenvalue \(\lambda\). (a) Show that \((A x, x)=(x, A x)\). Hint: See Problem 31 . (b) Show that \(\lambda(x, x)=\lambda(x, x)\), Hint: Recall that \(A x=\lambda x\). (c) Show that \(\lambda=\lambda\); that is, the cigenvalue \(\lambda\) is real.

Consider the system $$ x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\\ {-3} & {2} & {4}\end{array}\right) x $$ (a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector, namely, $$ \xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right) $$ (b) Using the information in part (a), write down one solution \(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the purely exponential form \(\mathbf{x}=\xi e^{y t}\). (c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2 t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{n}=\mathbf{\xi} $$ since \(\xi\) has already been found in part (a), solve the second equation for \(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i). (d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right) e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and \(\zeta\) satisfy the equations $$ (\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I}) \mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta} $$ The first two equations are the same as in part (c), so solve the third equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system (i). (e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i). (f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and third columns. Then find \(T^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A}\).

In each of Problems 9 through 14 find the general solution of the given system of equations. $$ \text { In each of Problems 9 through 14 find the general solution of the given system of equations. } $$

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that det \(\mathbf{A}=0\) and that \(y\) is a solution of \(\mathbf{A}^{*} \mathbf{y}=\mathbf{0} .\) Show that if \((\mathbf{b}, \mathbf{y})=0\) for every such \(\mathbf{y},\) then \(\mathbf{A} \mathbf{x}=\mathbf{b}\) has solutions. Note that the converse of Problem \(27 ;\) the form of the solution is given by Problem \(28 .\)
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