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lecture_41 - MA 265 LECTURE NOTES WEDNESDAY APRIL 23...

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MA 265 LECTURE NOTES: WEDNESDAY, APRIL 23 Systems with Diagonalizable Matrices Example. Consider the first-order homogeneous system dx dt = x ( t ) - y ( t ) dy dt = 2 x ( t ) + 4 y ( t ) We find the general solution. The idea is to make a change of variables to make things simpler. First, we compute the eigenvalues and eigenvectors of the coefficient matrix. Let’s write this system in the form d x dt = A x ( t ) where A = 1 - 1 2 4 and x ( t ) = x ( t ) y ( t ) . The characteristic polynomial of A is p ( λ ) = det [ λ I 2 - A ] = λ - 1 1 - 2 λ - 4 = λ 2 - 5 λ + 6 = ( λ - 2) ( λ - 3) so that the eigenvalues are λ 1 = 2 and λ 2 = 3. We see that the eigenvectors are A p 1 = λ 1 p 1 for p 1 = 1 - 1 ; A p 2 = λ 2 p 2 for p 2 = 1 - 2 . Define the following 2 × 2 matrix using these eigenvectors: P = 1 1 - 1 - 2 = D = P - 1 A P = 2 1 - 1 - 1 1 - 1 2 4 1 1 - 1 - 2 = 2 0 0 3 . Second, we make a change of variables to find a simpler first-order homogeneous system. Denote u ( t ) = P - 1 x ( t ) = u ( t ) = 2 x ( t ) + y ( t ) v ( t ) = - x ( t ) - y ( t ) We have the following differential equations: du dt = 2 dx dt + dy dt = 2 x ( t ) - y ( t ) + 2 x ( t ) + 4 y ( t ) = 4 x ( t ) + 2 y ( t ) = 2 u ( t ) dv dt = - dx dt - dy dt = - x ( t ) - y ( t ) - 2 x ( t ) + 4 y ( t ) = - 3 x ( t ) - 3 y ( t ) = 3 v ( t ) = d u dt = D u ( t ) . We saw in the previous lecture that the general solution to this simpler first-order system is the linear combination u ( t ) = b 1 u (1) ( t ) + b 2 u (2) ( t ) where u (1) ( t ) = 1 0 e
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