Science:Math Exam Resources/Courses/MATH307/December 2008/Question 08

MATH307 December 2008

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[hide] Question 08
Consider the linear differential equation ${\begin{aligned}x'(t)&=3x(t)+4y(t)\\y'(t)&=3x(t)+2y(t)\\\end{aligned}}$ with the initial conditions $\displaystyle x(0)=6$ , $\displaystyle y(0)=1$ . Find the solution $\displaystyle u(t)=(x(t),y(t))^{T}$ and determine the stability of the differential equation.

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[show] Hint 1
Start by writing the differential equation in matrix form u' = Au.

[show] Hint 2
If the eigenvalues of the corresponding matrix A are distinct, then the general solution of the linear differential equation is given by $\displaystyle u(t)=c_{1}v_{1}e^{\lambda _{1}t}+c_{2}v_{2}e^{\lambda _{2}t}$ where v₁ and v₂ are eigenvectors corresponding to the eigenvalues λ₁ and λ₂, respectively, and c₁, c₂ are constants.

[show] Hint 3
Next, use the initial condition to solve for the constants c₁ and c₂.

[show] Hint 4
Finally, the stability is determined by the sign of the (real part of the) eigenvalues.

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[show] Solution
To begin with we rewrite the system in matrix form: ${\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}}={\begin{bmatrix}3&4\\3&2\end{bmatrix}}{\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}$ and call this matrix A. We need the eigenvalues of A and hence find the zeros of the characteristic polynomials ${\begin{aligned}\det(A-\lambda I)&=\det \left({\begin{bmatrix}3-\lambda &4\\3&2-\lambda \end{bmatrix}}\right)\\&=(3-\lambda )(2-\lambda )-12\\&=\lambda ^{2}-5\lambda -6\\&=(\lambda -6)(\lambda +1)\end{aligned}}$ We see that λ₁ = 6 and λ₂ = -1. Next, we calculate corresponding eigenvectors. For λ₁ = 6 we find the eigenvectors in the null space of ${\begin{bmatrix}-3&4\\3&-4\end{bmatrix}}$ e.g. $v_{1}={\begin{bmatrix}4\\3\end{bmatrix}}$ . Similarly, for λ₂ = -1 we find the eigenvectors in the null space of ${\begin{bmatrix}4&4\\3&3\end{bmatrix}}$ e.g. $v_{2}={\begin{bmatrix}1\\-1\end{bmatrix}}$ . Hence, the general solution of the linear differential equation is given by ${\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}=c_{1}{\begin{bmatrix}4\\3\end{bmatrix}}e^{6t}+c_{2}{\begin{bmatrix}1\\-1\end{bmatrix}}e^{-t}$ To find the values of the constants c₁ and c₂ we plug in the values at the initial condition t = 0: ${\begin{aligned}x(0)&=6&=4c_{1}+c_{2}\\y(0)&=1&=3c_{1}-c_{2}\end{aligned}}$ which has the solution c₁ = 1, c₂ = 2. Therefore, the full solution of the linear differential equation with the given initial conditions is $u(t)={\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}={\begin{bmatrix}4\\3\end{bmatrix}}e^{6t}+{\begin{bmatrix}2\\-2\end{bmatrix}}e^{-t}$ Finally, since one of the eigenvalues is positive, the differential equation is unstable.