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Solve the initial Valve problem given as dx/dt = 2x+8y dy/dt = -x+y If x(0) = ( λ(0)/y(0)) = (2/-1)​

User Ben Hawker
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1 Answer

7 votes

I got you bro :)

We are given the system of differential equations:

dx/dt = 2x + 8y

dy/dt = -x + y

We can solve this system using matrix methods. First, we can rewrite the system in matrix form:

d/dt [x; y] = [2 8; -1 1] [x; y]

The matrix [2 8; -1 1] is called the coefficient matrix of the system.

Next, we can find the eigenvalues and eigenvectors of the coefficient matrix:

|2-λ 8| |-(λ-1) 8|

|-1 1-λ| = |-1 -(λ-1)|

Using the formula for the characteristic equation, we get:

(2-λ)(1-λ) + 8 = 0

Simplifying, we get:

λ^2 - 3λ + 10 = 0

Solving for λ using the quadratic formula, we get:

λ = (3 ± sqrt(29)i)/2

Since the eigenvalues have a non-zero imaginary part, the system has complex eigenvectors. The eigenvectors can be found using the formula:

(A - λI)v = 0

where A is the coefficient matrix, λ is an eigenvalue, and v is the eigenvector corresponding to λ. Solving this equation, we get:

v1 = (1 + sqrt(29)i)/4

v2 = (1 - sqrt(29)i)/4

Using the eigenvectors, we can diagonalize the coefficient matrix:

[2 8; -1 1] = [v1 v2] [λ1 0; 0 λ2] [v1 v2]^-1

where λ1 and λ2 are the eigenvalues and [v1 v2]^-1 is the inverse of the matrix [v1 v2].

Plugging in the values, we get:

[2 8; -1 1] = [(-sqrt(29)i/4) (sqrt(29)i/4)] [(3+isqrt(29))/4 0; 0 (3-isqrt(29))/4] [(-sqrt(29)i/4) (sqrt(29)i/4)]^-1

Simplifying, we get:

[2 8; -1 1] = [(sqrt(29)i/2) (sqrt(29)i/2)] [(1+isqrt(29))/2 0; 0 (1-isqrt(29))/2] [(sqrt(29)i/2) (sqrt(29)i/2)]^-1

Now we can solve the system of differential equations using the diagonalized matrix. We let:

[x; y] = [c1 v1; c2 v2]

where c1 and c2 are constants to be determined. Substituting into the matrix equation, we get:

d/dt [c1 v1; c2 v2] = [(1+isqrt(29))/2 0; 0 (1-isqrt(29))/2] [c1 v1; c2 v2]

Expanding, we get:

c1' v1 + c2' v2 = (1+isqrt(29))/2 c1 v1 + (1-isqrt(29))/2 c2 v2

c1' v1 + c2' v2 = (1+isqrt(29))/2 c1 v1 + (1-isqrt(29))/2 c2 v2

Solving for c1' and c2', we get:

c1' = (

User Ryan Tarpine
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