Paper 2, Section II, A

Differential Equations | Part IA, 2016

(a) By considering eigenvectors, find the general solution of the equations

dxdt=2x+5ydydt=x2y(†)\tag{†} \begin{aligned} &\frac{d x}{d t}=2 x+5 y \\ &\frac{d y}{d t}=-x-2 y \end{aligned}

and show that it can be written in the form

(xy)=α(5cost2costsint)+β(5sintcost2sint)\left(\begin{array}{l} x \\ y \end{array}\right)=\alpha\left(\begin{array}{c} 5 \cos t \\ -2 \cos t-\sin t \end{array}\right)+\beta\left(\begin{array}{c} 5 \sin t \\ \cos t-2 \sin t \end{array}\right)

where α\alpha and β\beta are constants.

(b) For any square matrix MM, exp(M)\exp (M) is defined by

exp(M)=n=0Mnn!\exp (M)=\sum_{n=0}^{\infty} \frac{M^{n}}{n !}

Show that if MM has constant elements, the vector equation dxdt=Mx\frac{d \mathbf{x}}{d t}=M \mathbf{x} has a solution x=exp(Mt)x0\mathbf{x}=\exp (M t) \mathbf{x}_{0}, where x0\mathbf{x}_{0} is a constant vector. Hence solve ()(†) and show that your solution is consistent with the result of part (a).

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