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Unformatted text preview: Factoring constant coefficient linear differential operators When we are working with constant coefficient differential operators, we can do algebra with the operators in much the same way as we can work with polynomials. If we define the operator D = d dt by the rule D ( y ) = dy dt = y Then we can build up our constant coefficient linear differential operators from D . If L ( y ) = ay 00 + by + cy then L ( y ) = ay 00 + by + cy = aD 2 y + bDy + cy = ( aD 2 + bD + c )( y ) so that we can write L = aD 2 + bD + c. We can also compose operators we have built up in this way, and the usual rules for multiplication (commutative, associative and distributive rule) all work in this new context. For example, we can compose the operators L 1 = D- r 1 and L 2 = D- r 2 and get the operator L 1 L 2 = ( D- r 1 )( D- r 2 ) = D 2- ( r 1 + r 2 ) D + r 1 r 2 = L because ( L 1 L 2 )( y ) = L 1 ( L 2 ( y )) = ( D- r 1 )(( D- r 2 ) y ) = ( D- r 1 )( y- r 2 y ) = (( y- r 2 y )- r 1 ( y- r 2 y )) = y 00- r 2 y- r 1 y + r 1 r 2 y = y 00- ( r 1 + r 2 ) y + r 1 r 2 y = L ( y ) . As with checking equality of functions, to check equality of operators, you must check that whatever argument (input function) you give the two oper- ators you compare, the output is the same function. This is another way of understanding the importance of the auxiliary equation. When find the roots and thus factor the auxiliary equation ar 2 + br + c = a ( r- r 1 )( r- r 2 ) 1 we are also factoring the differential operator L = aD 2 + bD + c = a ( D- r 1 )( D- r 2 ) = a ( D- r 2 )( D- r 1 ) (Notice that the order of the factors can be reversed). Since ( D- r )( e rt ) = 0, we are able to see again why y 1 = e r 1 t and y 2 = e r 2 t are solutions of the homogeneous equation L ( y ) = 0. If any factor annihilates a function, then the function will be a solution of the homogeneous equation. We can use this approach to try to produce a full complement of independent solutions. We can use this notation and way of thinking to redo the solution of the second order equation in the case where the root is repeated. In that case the equation we wish to solve is (( D- r ) 2 )( y ) = ( D- r )(( D- r )( y )) = 0 . We can use the factorization to solve this one order at a time. First we define u so that ( D- r )( y ) = u , then the equation we are trying to solve becomes ( D- r ) u = u- ru = 0 . This is linear with integrating factor e- rt (and separable) and the solution is u = C 1 e rt . Next we go back and solve for y in ( D- r )( y ) = y- ry = u = C 1 e rt . This is linear with integrating factor e- rt and solution y = e rt Z C 1 e rt e- rt dt = ( C 1 t + C 2 ) e rt . Of course we already knew that, but this is a nice way of looking at this that generalizes as we will see below....
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