Case 3: Repeated Eigenvalue

> lambda := %? ;

> V1 := vector( 2, [ %? , %? ] );

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

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The straight-line solution is:

> Y1 := exp( lambda *t) * convert( V1, matrix );

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To find a second linearly independent solution we must find any vector that satisfies or, equivalently, .

> sys := geneqns( evalm( A-lambda*I2 ), V2, V1 );

> soln := solve( sys, { x2, y2 } );

> V2 := subs( soln , x2=0, y2=0, evalm(V2) );

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

>