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Write the linear system of differential equations asA homogeneous power $t^{\lambda }\mathbf{v}$ solves this Cauchy-Euler differential system precisely when $A\mathbf{v}=\lambda \mathbf{v}$. The eigenvalues are $3$ and $6$, with eigenvectors $(2,1{)}^T$ and $(1,-1{)}^T$. The homogeneous solution is therefore $C{t}^3(2,1{)}^T+D{t}^6(1,-1{)}^T$.

For a particular solution, put ${\mathbf{z}}_p=t\mathbf{c}$. Then $(I-A)\mathbf{c}=\mathbf{b}$, which gives $\mathbf{c}=(3,0{)}^T$. At $t=1$, the two initial conditions become $2C+D+3=0$ and $C-D=0$, so $C=D=-1$. ThusThe two powers correspond to the two eigenvalues; the lower-degree particular term accounts for the constant forcing. Substitution gives both original right sides and the zero initial data.

The first and third equations form a constant-coefficient linear system of differential equationsThe eigenvectors $(1,1)$ and $(1,-1)$ have eigenvalues $2$ and $-4$. Applying the initial data givesThe remaining equation becomesAn integrating factor, or the usual exponential and trigonometric particular solutions, givesThese values satisfy all three initial conditions.

The linearized vortensity amplitude isSince the exact vortensity obeys $D_t\zeta =0$ and the time-dependent $k_x$ cancels the background shear in the phase, linearization gives

On the zero-vortensity branch,Substitution in the two momentum equations yields the linear system of differential equationswithwhere $k_x=k_x(t)$ is the result from part (ii).