24 Solving nonhomogeneous systems
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1 4 Solving nonhomogeneous sysems Consider nonhomogeneous sysem ẏ Ay f(), A [a ij n n, f : R R n () Similarly o he case of linear ODE of he n-h order, i is rue ha Proposiion The general soluion o sysem () is given by he sum of he general soluion o he homogeneous sysem plus a paricular soluion o he nonhomogeneous one: y() y h () y p () The proof is lef an an exercise and relies on he fac ha if y and y solve () hen y y solves homogeneous sysem Therefore, o solve sysem () we need somehow find a paricular soluion o he nonhomogeneous sysem and use he echnique from he previous lecures o obain soluion o he homogeneous sysem Recall ha for he linear equaions we consider hree approaches o solve nonhomogeneous equaions: variaion of he consan (or variaion of he parameer), mehod of an educaed guess and he Laplace ransform mehod Similarly, we can use he same mehods here I will sar wih he mos imporan heoreically mehod: variaion of he consans 4 Variaion of he consans Le Φ() be a fundamenal marix soluion o he homogeneous sysem ẏ Ay () Hence, he general soluion can be wrien as y h () Φ()c, where c (C,, C n ) is a vecor of arbirary consans Assume ha his vecor is no consan, bu a funcion of : and plug y() Φ()c() ino () We find and since Φ() AΦ(), hen, finally, c c(), Φ()c() ċ()φ() AΦ()c() f(), ċ() Φ ()f(), which has he soluion c() c Φ (τ)f(τ) dτ, where now c is he vecor of arbirary consans Using his soluion, we obain y() Φ()c }{{} Φ() Φ (τ)f(τ) dτ y h () }{{} y p () MATH66: Inro o ODE by Arem Novozhilov, aremnovozhilov@ndsuedu Fall 3
2 This formula is very convenien heoreically, bu in acual calculaions usually requires quie a few seps Since he marix exponen is a special fundamenal marix, we can rewrie he las soluion in he form y() e A c e A( τ) f(τ) dτ This las formula requires only a sligh modificaion if we are given he iniial condiions y( ) y : y() e A( ) y e A( τ) f(τ) dτ Example Solve ẏ [ 3 4 y [ e, y() e [ Eigenvalues and eigenvecors We have [ 3 λ 4 de(a λi) de λ λ (λ ), λ hence we have one eigenvalue λ mulipliciy wo Consider [ [ [ 4 v, v which implies ha he eigenvalue λ has only one eigenvecor v (, ), and hence our firs paricular soluion o he homogeneous sysem can be wrien as [ y () e To find he second linearly independen soluion, we will need o look for he generalized eigenvecor ha solves (A λi) v We find ha (A λi) [, herefore any vecor can be aken as a generalized eigenvecor, however, we need o remember ha we need o ake one, which is linearly independen of v I choose o ake [ v Therefore, he second paricular soluion o he linear sysem is y () e ( I (A λi) ) ([ [ ) v e 4 v [ e
3 Fundamenal marix soluion and marix exponen We found ha [ e ( )e Φ() e e is a fundamenal marix soluion for he corresponding linear sysem We also have ha [ Φ(), herefore Φ () [, and hence [ e A Φ()Φ e ( )e () [ e e Solving nonhomogeneous sysem We have [ e Aτ τ 4τ e τ, τ τ herefore, Finally, hence y() e A y 4 Mehod of an educaed guess [ e Aτ f(τ) dτ [ e A [ ( ) [ ( )e 4e e ( )e e, [ [ e A( τ) f(τ) dτ e ( ) e e [ Someimes i is possible o guess wha is he form of a paricular soluion Consider, eg, sysem of he form ẏ Ay we a, where w is a consan vecor, and a is a consan Le us look for a paricular soluion in he form We find ha y p () xe a ax Ax w (A ai)x w Assuming ha a is no an eigenvalue of marix A, we can solve he las sysem x (A ai) w 3
4 Example 3 Solve ẏ [ y 3 [ e 3 As usual we find ha marix A has eigenvalues λ and λ wih he eigenvecors v (, ) and v (3, ), herefore, he general soluion o he homogeneous sysem is [ [ y h () C e 3 C e, where C, C are arbirary consans Since 3 is no an eigenvalue of A, we can look for a paricular soluion o he nonhomogeneous sysem in he form [ A y p () e 3, B where A and B are consan o be deermined Afer plugging y p () and canceling all he exponens, we obain he sysem 3A A B, 3B 3A B, which has he unique soluion A /, B / Therefore, he general soluion o our sysem is given by [ [ y() C e 3 C e [ e 3 43 Using he Laplace ransform We can also use he Laplace ransform o aack nonhomogeneous problems Assume ha we need o solve sysem () and le Y (s) be he Laplace ransform of y(): We also will need he fac ha Y (s) L {y()} L {ẏ()} sy (s) y Now by applying he Laplace ransform o he lef and righ hand sides of (), we find or sy (s) y AY (s) F (s), Y (s) (si A) (y F (s)), which gives us he formal soluion o he problems, provided ha we are able o find he inverse Laplace ransform y() L {Y (s)} 4
5 Example 4 Consider again ẏ [ 3 4 y [ e, y() e [ By applying he Laplace ransform, we find [ s 3 4 Y s [ [ s, s from where he vecor Y (s) can be found (using any mehod you prefer o solve he linear sysem) as Y (s) [ s 3 s 3 (s ) (s ) 3 s s (s ) (s ) 3 Using parial fracion decomposiion, we find, eg, for he firs elemen ha s 4s 3 s 3 (s ) 3 s (s ) (s ) 3, herefore, and similarly for he second elemen L { } e e e, L { } e e Therefore, he final answer is, as before, [ y() e 5
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