Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 8A (Differential Equations)

Write the system as ${\mathbf{x}}^{\prime }=M\mathbf{x}$, with $M=\left(\begin{array}{cc}2& 5\\ -1& -2\end{array}\right)$. Its characteristic polynomial is ${\lambda }^2+1$, so its eigenvalues are $i$ and $-i$. An eigenvector for $i$ is $v=(5,-2+i{)}^{\mathsf{T}}$. Thus $e^{it}v$ is a complex solution, and its real part and imaginary part give real solutions:Their initial vectors are $(5,-2{)}^{\mathsf{T}}$ and $(0,1{)}^{\mathsf{T}}$, with determinant $5\ne 0$. Consequently they are linearly independent and span the two-dimensional solution space.

For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.