Math Problem Statement

Use the elimination method to find a general solution for the given linear​ system, where differentiation is with respect to t. StartFraction dx Over dt EndFraction equals 5 x minus y StartFraction dy Over dt EndFraction equals 4 x plus 5 y Question content area bottom Part 1 Eliminate x and solve the remaining differential equation for y. Choose the correct answer below. A. ​y(t)equalsUpper C 1 e Superscript negative 5 t Baseline plus Upper C 2 t e Superscript negative 5 t B. ​y(t)equalsUpper C 1 e Superscript negative 5 t Baseline cosine 2 t plus Upper C 2 e Superscript negative 5 t Baseline sine 2 t C. ​y(t)equalsUpper C 1 e Superscript 5 t Baseline plus Upper C 2 t e Superscript 5 t D. ​y(t)equalsUpper C 1 e Superscript 5 t Baseline cosine 2 t plus Upper C 2 e Superscript 5 t Baseline sine 2 t E. The system is degenerate.

Solution

To solve this system of linear differential equations using the elimination method, we will:

  1. Write the system of equations: dxdt=5xy\frac{dx}{dt} = 5x - y dydt=4x+5y\frac{dy}{dt} = 4x + 5y

  2. Differentiate the first equation with respect to tt: d2xdt2=ddt(5xy)=5dxdtdydt.\frac{d^2x}{dt^2} = \frac{d}{dt}(5x - y) = 5\frac{dx}{dt} - \frac{dy}{dt}. Substitute dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} from the original system: d2xdt2=5(5xy)(4x+5y).\frac{d^2x}{dt^2} = 5(5x - y) - (4x + 5y). Simplify: d2xdt2=25x5y4x5y=21x10y.\frac{d^2x}{dt^2} = 25x - 5y - 4x - 5y = 21x - 10y.

  3. Eliminate yy using the first equation: From dxdt=5xy\frac{dx}{dt} = 5x - y, solve for yy: y=5xdxdt.y = 5x - \frac{dx}{dt}. Substitute yy into the second-order equation: d2xdt2=21x10(5xdxdt).\frac{d^2x}{dt^2} = 21x - 10(5x - \frac{dx}{dt}). Simplify: d2xdt2=21x50x+10dxdt.\frac{d^2x}{dt^2} = 21x - 50x + 10\frac{dx}{dt}. d2xdt210dxdt+29x=0.\frac{d^2x}{dt^2} - 10\frac{dx}{dt} + 29x = 0.

  4. Solve the characteristic equation: The characteristic equation is: r210r+29=0.r^2 - 10r + 29 = 0. Solve using the quadratic formula: r=(10)±(10)24(1)(29)2(1)=10±1001162.r = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(29)}}{2(1)} = \frac{10 \pm \sqrt{100 - 116}}{2}. r=10±162=10±4i2.r = \frac{10 \pm \sqrt{-16}}{2} = \frac{10 \pm 4i}{2}. r=5±2i.r = 5 \pm 2i.

  5. Write the general solution for x(t)x(t): The solution is: x(t)=e5t(C1cos(2t)+C2sin(2t)).x(t) = e^{5t}(C_1\cos(2t) + C_2\sin(2t)).

  6. Solve for y(t)y(t): Substitute x(t)x(t) into y=5xdxdty = 5x - \frac{dx}{dt}. After calculations, the solution for y(t)y(t) is: y(t)=e5t(C1cos(2t)+C2sin(2t)).y(t) = e^{5t}(C_1\cos(2t) + C_2\sin(2t)).

The correct answer is: D. y(t)=C1e5tcos(2t)+C2e5tsin(2t)y(t) = C_1 e^{5t} \cos(2t) + C_2 e^{5t} \sin(2t).

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Related Questions:

  1. How can you verify the solution of a system of differential equations?
  2. What does the characteristic equation tell us about the behavior of the solutions?
  3. Why do we see oscillatory behavior in solutions involving complex eigenvalues?
  4. How does the elimination method differ from matrix-based methods for solving linear systems?
  5. What physical systems might correspond to this type of differential equation?

Tip:

For second-order differential equations with complex roots, the solution always includes exponential growth/decay combined with sinusoidal functions.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Elimination Method
Complex Roots
Characteristic Equations

Formulas

dx/dt = 5x - y
dy/dt = 4x + 5y
Characteristic equation: r^2 - 10r + 29 = 0
General solution for second-order equations with complex roots: e^(αt)(C1*cos(βt) + C2*sin(βt))

Theorems

Superposition principle for linear differential equations
Quadratic formula for roots: r = [-b ± sqrt(b^2 - 4ac)] / 2a

Suitable Grade Level

Undergraduate Mathematics or Advanced High School Calculus