Case 3: Repeated Eigenvalue

> lambda := ? ;

> V1 := matrix( 2, 1, [ ? , ? ] );

> V2 := matrix( 2, 1, [ x2, y2 ] );

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The straight-line solution is:

> Y1 := exp( lambda *t) * evalm( V1 );

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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) );

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> Y2 := exp( lambda *t) * ( t * evalm( V1 ) + evalm( V2 ) );

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