Just as there are simultaneous algebraic equations (where a pair of numbers have to satisfy a pair of equations) there are systems of differential equations, (where a pair of functions have to satisfy a pair of differential equations).Indicate which pairs of functions satisfy this system. It will take some time to make all of the calculations.
y_1' = y_1 -2 y_2 \qquad y_2' = 3y_1 - 4 y_2


A. y_1 = \sin(x) +\cos(x) \qquad y_2 = \cos(x) - \sin(x)
B. y_1 = \sin(x) \qquad y_2 = \cos(x)
C. y_1 = \cos(x) \qquad y_2 = -\sin(x)
D. y_1 = e^{-x} \qquad y_2=e^{-x}
E. y_1 = e^x \qquad y_2=e^x
F. y_1 = e^{4x} \qquad y_2 = e^{4x}
G. y_1 = 2e^{-2x} \qquad y_2 = 3e^{-2x}


As you can see, finding all of the solutions, particularly of a system of equations, can be complicated and time consuming. It helps greatly if we study the structure of the family of solutions to the equations. Then if we find a few solutions we will be able to predict the rest of the solutions using the structure of the family of solutions.

Answers

Answer 1
Answer:

Answer: D and G.

Step-by-step explanation:

For options D and G we will show that both differential equations are satisfied. For the other options we will show the pairs don't solve one of the equations.

  • A. y_1 '= \cos(x)-\sin x and y_1-2y_2= \sin x+\cos x -2(\cosx -\sin x )=3\sin x- \cos x \neq \cos x-\sin x (when x=0 the left side is -1 and the right side is 1) so the equation y_1'=y_1 - 2y_2 is not satisfied.
  • B. y_2 '= -\sin x and 3y_1-4y_2= 3\sin x-4\cos x \neq -\sin x so the equation y_2'=3y_1-4y_2 is not satisfied.
  • C. y_1 '= -\sin(x) and y_1-2y_2= \cos x -2\sin x \neq -\sin x so these pairs don't solve the equation y_1'=y_1-2y_2.
  • D. Since y_1=y_2=e^(-x) then y_1'=y_2'=-e^(-x). The first equation is satisfied, because y_1-2y_2=e^(-x)-2e^(-x)=-e^(-x)=y_1'. The second equation is also satisfied: 3y_1-4y_2=3e^(-x)-4e^(-x)=-e^(-x)=y_2'.
  • E. y_2'=e^x and 3y_1-4y_2= 3e^x-4e^x=-e^x\neq -e^x so they don't satisfy the equation y_2'=3y_1-4y_2.
  • F.  y_1 '= 4e^(4x) and y_1-2y_2= e^(4x)-2e^(4x)=-e^(4x) \neq 4e^(4x), then the equation y_1'=y_1-2y_2 is not satisfied.
  • G. In this case, y_1=2e^(-2x) and y_2=3e^(-2x). Computing derivatives, y_1'=-4e^(-2x) and y_2'=-6e^(-2x). The first equation is satisfied, because y_1-2y_2=2e^(-2x)-6e^(-2x) =-4e^(-2x)=y_1'. The second equation is also satisfied: 3y_1-4y_2= 6e^(-2x)-12e^(-2x)=-6e^(-2x)=y_2'.
Answer 2
Answer:

Final answer:

The pairs of functions that satisfy the given system of differential equations are Option D (y_1 = e^(-x), y_2 = e^(-x)) and Option E (y_1 = e^x, y_2 = e^x).

Explanation:

The given system of differential equations is:

y_1' = y_1 - 2y_2

y_2' = 3y_1 - 4y_2

To determine which pairs of functions satisfy this system, we can substitute each option into the system and check if they satisfy the equations.

Let's go through each option:

  1. Option A: y_1 = sin(x) + cos(x), y_2 = cos(x) - sin(x)
    By substituting these functions into the system, we get:
    y_1' = cos(x) - sin(x) - 2(cos(x) - sin(x)) = -sin(x) - 4cos(x)
    y_2' = sin(x) + cos(x) - 4(cos(x) - sin(x)) = 5sin(x) - 3cos(x)
    These functions do not satisfy the system of differential equations.
  2. Option B: y_1 = sin(x), y_2 = cos(x)
    By substituting these functions into the system, we get:
    y_1' = cos(x) - 2cos(x) = -cos(x)
    y_2' = 3sin(x) - 4cos(x)
    These functions do not satisfy the system of differential equations.
  3. Option C: y_1 = cos(x), y_2 = -sin(x)
    By substituting these functions into the system, we get:
    y_1' = -sin(x) + 2sin(x) = sin(x)
    y_2' = 3cos(x) - 4(-sin(x)) = 3cos(x) + 4sin(x)
    These functions do not satisfy the system of differential equations.
  4. Option D: y_1 = e^(-x), y_2 = e^(-x)
    By substituting these functions into the system, we get:
    y_1' = -e^(-x) - 2e^(-x) = -3e^(-x)
    y_2' = 3e^(-x) - 4e^(-x) = -e^(-x)
    These functions satisfy the system of differential equations.
  5. Option E: y_1 = e^x, y_2 = e^x
    By substituting these functions into the system, we get:
    y_1' = e^x - 2e^x = -e^x
    y_2' = 3e^x - 4e^x = -e^x
    These functions satisfy the system of differential equations.
  6. Option F: y_1 = e^(4x), y_2 = e^(4x)
    By substituting these functions into the system, we get:
    y_1' = 4e^(4x) - 2e^(4x) = 2e^(4x)
    y_2' = 3e^(4x) - 4e^(4x) = -e^(4x)
    These functions do not satisfy the system of differential equations.
  7. Option G: y_1 = 2e^(-2x), y_2 = 3e^(-2x)
    By substituting these functions into the system, we get:
    y_1' = -2e^(-2x) - 2(3e^(-2x)) = -8e^(-2x)
    y_2' = 3(2e^(-2x)) - 4(3e^(-2x)) = -6e^(-2x)
    These functions satisfy the system of differential equations.

Therefore, the pairs of functions that satisfy the given system of differential equations are Option D (y_1 = e^(-x), y_2 = e^(-x)) and Option E (y_1 = e^x, y_2 = e^x).

Learn more about Systems of Differential Equations here:

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Answers

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Answers

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Answers

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