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Find the general solution of the systems. \(x^{\prime}=4 x-5 y+4 z\) \(y^{\prime}=-y+4 z\) \(z^{\prime}=z\)

Short Answer

Expert verified
The general solution is: \[ x = W t^2 e^t + V t e^t + Qe^t, \] \[ y = 4 C t e^t + D e^t, \] \[ z = C e^t, \] with constants W, V, Q, C, D.

Step by step solution

01

Identify the System of Equations

We are given a system of three linear differential equations: \[ x^{\prime}=4x-5y+4z \]\[ y^{\prime}=-y+4z \]\[ z^{\prime}=z \]Notice that the equations involve derivatives of \( x \), \( y \), and \( z \) with respect to time \( t \).
02

Solve the Third Equation

The equation \( z^{\prime} = z \) is a simple first-order linear differential equation. Its general solution is:\[ z = Ce^t \]where \( C \) is a constant.
03

Substitute the Solution for z into the Second Equation

Substitute \( z = Ce^t \) into the second equation:\[ y^{\prime} = -y + 4Ce^t \]This is a non-homogeneous first-order linear differential equation in \( y \).
04

Solve the Differential Equation for y

The differential equation \( y^{\prime} = -y + 4Ce^t \) can be solved using an integrating factor. The integrating factor \( \mu(t) = e^{\int -1 \, dt} = e^{-t} \). Multiply the entire equation by \( e^{-t} \):\[ e^{-t}y^{\prime} + e^{-t}y = 4Ce^0 \]This becomes:\[ \frac{d}{dt}(e^{-t}y) = 4C \]Integrating both sides:\[ e^{-t}y = 4Ct + D \]Thus, the solution for \( y \) is:\[ y = 4Ct e^t + De^t \]where \( D \) is a constant.
05

Substitute Solutions for y and z into the First Equation

Now substitute \( z = Ce^t \) and \( y = 4Ct e^t + De^t \) into the first equation:\[ x^{\prime} = 4x - 5(4Ct e^t + De^t) + 4Ce^t \]This simplifies to:\[ x^{\prime} = 4x - 20Ct e^t - 5De^t + 4Ce^t \]Simplified:\[ x^{\prime} = 4x - 20Ct e^t - 5De^t + 4Ce^t \].
06

Solve the Differential Equation for x

This is a non-homogeneous differential equation in \( x \). Suppose \( x = Ve^t + Wte^t \).We differentiate and substitute to form:\[ Ve^t + Wte^t + Wx = 4(Ve^t + Wte^t) - 20Ct e^t - 5De^t + 4Ce^t \]Solve using an analogous integrating factor.
07

Assemble the General Solution

Once we solve for \( V \) and \( W \) in terms of constants, we plug these back to form:\[ x = Wt^2e^t + Vte^t + Qe^t \] where \( Q \) is another constant. We combine this with earlier solutions for \( y \) and \( z \) for the complete general solution:\[ x = Wt^2 e^{t} + Vt e^{t} + Qe^t \]\[ y = 4Ct e^t + De^t \]\[ z = Ce^t \].

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Key Concepts

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

General Solution
When tackling a system of linear differential equations, we aim to find the general solution. This accounts for all possible solutions to the equations, including arbitrary constants.
The general solution encompasses both the homogeneous solution, where the system equals zero, and a particular solution that fits the non-homogeneous part. In essence,
  • Homogeneous solutions are usually derived first, often using techniques such as characteristic equations or matrix eigenvalues.
  • Particular solutions employ methods like undetermined coefficients or variation of parameters for non-zero terms.
In our exercise, the system involved three interconnected equations. Step by step, we worked through each, starting from simpler to more complex. We derived individual solutions for each variable which, when combined, form the overall general solution.
The general solution, reflecting the entire behavior of the system, integrates constants like \( C \), \( D \), \( V \), and \( W \), which adjust to specific initial conditions.
First-Order Differential Equations
A first-order differential equation involves the first derivative of the function but no higher derivatives. It is fundamental in modeling various real-world scenarios, from physics to economics.
Solving a first-order linear differential equation usually involves finding an integrating factor or using separation of variables:
  • The integrating factor is a function multiplied by the equation to make it easier to solve, particularly by simplifying the expression to allow integration.
  • Once transformed, the equation can be integrated directly leading us to the solution for the dependent variable.
In the given problem, the equations for \( z \) and \( y \) are first-order. We solved these by leveraging their structure:
For \( z' = z \), the simplicity allowed for immediate recognition of the solution form \( z = Ce^t \).
For \( y' = -y + 4Ce^t \), an integrating factor was instrumental in solving, resulting in \( y = 4Ct e^t + De^t \), illustrating the interconnectedness of \( y \) and \( z \).
System of Equations
A system of equations consists of multiple equations that share variables and need to be solved simultaneously. With linear differential systems:
  • We often encounter situations where one solution depends on others, showing complex interaction between equations.
  • Solving begins by isolating, then substituting solutions back to progressively reduce complexity.
To illustrate, our exercise had three variables: \( x \), \( y \), and \( z \). By starting with the equation \( z' = z \), and moving onto \( y \), we laid the groundwork to address \( x' \) last.
This strategy is effective because solving simpler equations first often helps in reducing the remaining equations’ complexity, offering partial solutions as useful substitutions.
Each variable's equation is systematically addressed, utilizing previous solutions to streamline the process and unify all three equations into a coherent and comprehensive general solution.

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Most popular questions from this chapter

Find the general solution of the given system. Write your solution in the form $$ \mathbf{y}(t)=e^{\lambda t}\left[\left(C_{1}+C_{2} t\right) \mathbf{v}_{1}+C_{2} \mathbf{v}_{2}\right], $$ where \(\mathbf{v}_{1}\) is an eigenvector and \(\mathbf{v}_{2}\) satisfies \((A-\lambda I) \mathbf{v}_{2}=\mathbf{v}_{1}\). Without the use of a computer or a calculator, sketch the half-line solutions. Sketch exactly one solution in each region separated by the half-line solutions. Use a numerical solver to verify your result when finished. Hint: The solutions in this case want desperately to spiral but are prevented from doing so by the half-line solutions (solutions cannot cross). However, the suggestions regarding clockwise or counterclockwise rotation in the subsection on spiral sources apply nicely in this situation. \(\mathbf{y}^{\prime}=\left(\begin{array}{rr}6 & 4 \\ -1 & 2\end{array}\right) \mathbf{y}\)

Find the eigenvalues and their associated eigenvectors. State the algebraic and geometric multiplicity of each eigenvalue. In the case where there are enough independent eigenvectors, state the general solution of the given system. \(x^{\prime}=x\) \(y^{\prime}=x+y\) \(z^{\prime}=-10 x+8 y+5 z\)

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. \(\mathbf{y}(t)=C_{1} e^{3 t}\left(\begin{array}{l}4 \\\ 1\end{array}\right)+C_{2} e^{t}\left(\begin{array}{l}1 \\\ 5\end{array}\right)\)

Each of the matrices has only one eigenvalue \(\lambda\). In each exercise, determine the smallest \(k\) such that \((A-\lambda I)^{k}=0\). The use the fact that $$ e^{t A}=e^{\lambda t}\left[I+t(A-\lambda I)+\frac{t^{2}}{2 !}(A-\lambda I)^{2}+\cdots\right]$$ to compute \(e^{t A}\). \(A=\left(\begin{array}{rrr}-2 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & -4 & -4\end{array}\right)\)

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. \(A=\left(\begin{array}{rrrrrr}0 & -30 & -42 & 40 & -48 & 14 \\ 1 & 7 & 9 & -9 & 10 & -2 \\ -1 & 5 & 8 & -6 & 6 & -2 \\ 2 & 45 & 64 & -60 & 72 & -20 \\ 2 & 33 & 47 & -45 & 55 & -15 \\ 0 & 7 & 11 & -10 & 10 & -1\end{array}\right)\)

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