Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 7A (Differential Equations)

Write the coefficient matrix as $M$. Its characteristic polynomial is $\det (rI-M)=r(r-2)(r+2)$. Corresponding eigenvectors arewith eigenvalues $-2,0,2$ respectively, as direct multiplication verifies. Distinct eigenvalues make these vectors a basis. Expanding a solution as $z_{-}{v}_{-}+z_0{v}_0+z_{+}{v}_{+}$ decouples the linear ordinary differential equations into ${\dot{z}}_{-}=-2z_{-}$, ${\dot{z}}_0=0$, ${\dot{z}}_{+}=2z_{+}$. Thus the general solution is

The forcing vector has the eigenvector decompositionThere is no component in the zero eigenvalue direction. The scalar inhomogeneous linear differential equations in this basis areFor the first, a particular solution is $(\lambda +1)e^{2t}/4$. For the third, multiplying by the integrating factor $e^{-2t}$ gives $(e^{-2t}{z}_{+}{)}^{\prime }=\lambda -1$. Consequently for every real $\lambda$When $\lambda =1$, the resonant component vanishes and a particular solution is $\frac{1}{2}{v}_{-}{e}^{2t}$, of the stipulated constant-vector form. If $\lambda \ne 1$, substituting $c{e}^{2t}$ would require $(2I-M)c=2(-\lambda ,1,\lambda {)}^T$. In the eigenbasis the left side has zero $v_{+}$ component, while the right side has component $\lambda -1$. This is impossible. The factor $t$ is therefore unavoidable: it is resonance in a differential equation.

At $\lambda =-1$ the $v_{-}$ forcing vanishes and the $v_{+}$ forcing coefficient is $-2$. The initial vector decomposes asThus $A=1/2$, $B=0$, $C=-1/2$ in part (b), and the requested solution isEquivalently,These expressions have the required initial values and solve all three linear ordinary differential equations.