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1. (20%)\mathbf{( 2 0 \% )} Find the general solution of the given system (a) $\qu...
Jun 19, 2024
Solution by Steps
step 1
We need to find the general solution of the system of differential equations given by X′=(1amp;1amp;40amp;2amp;01amp;1amp;1)X \mathbf{X}^{\prime} = \left( \begin{array}{ccc} 1 & 1 & 4 \\ 0 & 2 & 0 \\ 1 & 1 & 1 \end{array} \right) \mathbf{X} with initial condition X(0)=(130) \mathbf{X}(0) = \left( \begin{array}{c} 1 \\ 3 \\ 0 \end{array} \right)
step 2
First, we find the eigenvalues of the matrix A=(1amp;1amp;40amp;2amp;01amp;1amp;1) A = \left( \begin{array}{ccc} 1 & 1 & 4 \\ 0 & 2 & 0 \\ 1 & 1 & 1 \end{array} \right) . The characteristic polynomial is det⁡(A−λI)=0 \det(A - \lambda I) = 0
step 3
Solving det⁡(1−λamp;1amp;40amp;2−λamp;01amp;1amp;1−λ)=0 \det \left( \begin{array}{ccc} 1 - \lambda & 1 & 4 \\ 0 & 2 - \lambda & 0 \\ 1 & 1 & 1 - \lambda \end{array} \right) = 0 , we get the eigenvalues λ1=3 \lambda_1 = 3 , λ2=2 \lambda_2 = 2 , and λ3=−1 \lambda_3 = -1
step 4
Next, we find the eigenvectors corresponding to each eigenvalue. For λ1=3 \lambda_1 = 3 , solving (A−3I)v=0 (A - 3I) \mathbf{v} = 0 , we get v1=(201) \mathbf{v}_1 = \left( \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right)
step 5
For λ2=2 \lambda_2 = 2 , solving (A−2I)v=0 (A - 2I) \mathbf{v} = 0 , we get v2=(5−32) \mathbf{v}_2 = \left( \begin{array}{c} 5 \\ -3 \\ 2 \end{array} \right)
step 6
For λ3=−1 \lambda_3 = -1 , solving (A+I)v=0 (A + I) \mathbf{v} = 0 , we get v3=(−201) \mathbf{v}_3 = \left( \begin{array}{c} -2 \\ 0 \\ 1 \end{array} \right)
step 7
The general solution is X(t)=c1e3tv1+c2e2tv2+c3e−tv3 \mathbf{X}(t) = c_1 e^{3t} \mathbf{v}_1 + c_2 e^{2t} \mathbf{v}_2 + c_3 e^{-t} \mathbf{v}_3
step 8
Using the initial condition X(0)=(130) \mathbf{X}(0) = \left( \begin{array}{c} 1 \\ 3 \\ 0 \end{array} \right) , we solve for c1,c2,c3 c_1, c_2, c_3
step 9
We get c1=1 c_1 = 1 , c2=0 c_2 = 0 , and c3=3 c_3 = 3 . Thus, the solution is X(t)=e3t(201)+3e−t(−201) \mathbf{X}(t) = e^{3t} \left( \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right) + 3 e^{-t} \left( \begin{array}{c} -2 \\ 0 \\ 1 \end{array} \right)
Question 1(b)
step 1
We need to solve the system of differential equations dxdt=−2x+2y \frac{d x}{d t} = -2x + 2y and dydt=x+3y \frac{d y}{d t} = x + 3y
step 2
Write the system in matrix form: X′=(−2amp;21amp;3)X \mathbf{X}^{\prime} = \left( \begin{array}{cc} -2 & 2 \\ 1 & 3 \end{array} \right) \mathbf{X}
step 3
Find the eigenvalues of the matrix A=(−2amp;21amp;3) A = \left( \begin{array}{cc} -2 & 2 \\ 1 & 3 \end{array} \right) . The characteristic polynomial is det⁡(A−λI)=0 \det(A - \lambda I) = 0
step 4
Solving det⁡(−2−λamp;21amp;3−λ)=0 \det \left( \begin{array}{cc} -2 - \lambda & 2 \\ 1 & 3 - \lambda \end{array} \right) = 0 , we get the eigenvalues λ1=1 \lambda_1 = 1 and λ2=−4 \lambda_2 = -4
step 5
Find the eigenvectors corresponding to each eigenvalue. For λ1=1 \lambda_1 = 1 , solving (A−I)v=0 (A - I) \mathbf{v} = 0 , we get v1=(21) \mathbf{v}_1 = \left( \begin{array}{c} 2 \\ 1 \end{array} \right)
step 6
For λ2=−4 \lambda_2 = -4 , solving (A+4I)v=0 (A + 4I) \mathbf{v} = 0 , we get v2=(1−1) \mathbf{v}_2 = \left( \begin{array}{c} 1 \\ -1 \end{array} \right)
step 7
The general solution is X(t)=c1etv1+c2e−4tv2 \mathbf{X}(t) = c_1 e^{t} \mathbf{v}_1 + c_2 e^{-4t} \mathbf{v}_2
step 8
Using the initial condition, solve for c1 c_1 and c2 c_2
Answer
The general solution for part (a) is X(t)=e3t(201)+3e−t(−201) \mathbf{X}(t) = e^{3t} \left( \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right) + 3 e^{-t} \left( \begin{array}{c} -2 \\ 0 \\ 1 \end{array} \right) . For part (b), the general solution is X(t)=c1et(21)+c2e−4t(1−1) \mathbf{X}(t) = c_1 e^{t} \left( \begin{array}{c} 2 \\ 1 \end{array} \right) + c_2 e^{-4t} \left( \begin{array}{c} 1 \\ -1 \end{array} \right) .
Key Concept
Eigenvalues and Eigenvectors
Explanation
The solution to a system of linear differential equations can be found using the eigenvalues and eigenvectors of the coefficient matrix.
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