Case 3: Repeated Eigenvalue
> lambda := ? ;
> V1 := matrix( 2, 1, [ ? , ? ] );
> V2 := matrix( 2, 1, [ x2, y2 ] );
>
The straight-line solution is:
> Y1 := exp( lambda *t) * evalm( V1 );
>
To find a second linearly independent solution we must find any vector that satisfies or, equivalently, .
> eqn := evalm( A &* V2 - lambda *V2 = V1 );
> soln := solve( { ? *x2 + ? *y2 = ? }, { x2, y2 } );
> V2 := subs( soln , y2=1, evalm(V2) );
>
> Y2 := exp( lambda *t) * ( t * evalm( V1 ) + evalm( V2 ) );
>