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# ch7 - CHAPTER 7 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS...

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Section 7.1 333 CHAPTER 7 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS SECTION 7.1 FIRST-ORDER SYSTEMS AND APPLICATIONS 1. Let 2 1 2 1 2 and , so 7 3 . x x x x x x x x x t ′′ = = = = = − + Equivalent system: 1 x = x 2 , 2 x = - 7 x 1 - 3 x 2 + t 2 2. Let 1 2 1 3 2 4 3 , , , , so x x x x x x x x x x x ′′ ′′′ = = = = = = = (4) 4 3 6 cos3 . x x x x x t ′′ = = + + Equivalent system: 1 x = x 2 , 2 x = x 3 , 3 x = x 4 , 4 x = - x 1 + 3 x 2 - 6 x 3 + cos 3 t 3. Let ( ) 2 2 1 2 1 2 and , so 1 / . x x x x x x x t x tx t ′′ = = = = = Equivalent system: 1 x = x 2 , t 2 2 x = (1 - t 2 ) x 1 - tx 2 4. Let ( ) 2 3 1 2 1 3 2 3 , , , so 5 3 2 ln / x x x x x x x x x x x tx t x t t ′′ ′′′ ′′ = = = = = = = − + . Equivalent system: 1 x = x 2 , 2 x = x 3 , t 3 3 x = - 5 x 1 - 3 tx 2 + 2 t 2 x 3 + ln t 5. Let ( ) 2 1 2 1 3 2 3 , , , so cos . x x x x x x x x x x x x ′′ ′′′ = = = = = = = + Equivalent system: 1 x = x 2 , 2 x = x 3 , 3 x = x 2 2 + cos x 1 6. Let 1 2 1 1 2 1 2 2 , , , so 5 4 , 4 5 . x x x x x y y y y y x x x y y y x y ′′ ′′ = = = = = = = = = = − + Equivalent system:

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334 Chapter 7 1 x = x 2 , 2 x = 5 x 1 - 4 y 1 1 y = y 2 , 2 y = - 4 x 1 + 5 y 1 7. Let 2 2 3/ 2 1 2 1 1 2 1 2 , , , so /( ) , x x x x x y y y y y x x kx x y ′′ = = = = = = = = − + 2 2 3/ 2 2 /( ) . y y ky x y ′′ = = − + Equivalent system: 1 x = x 2 , 2 x = ( ) 3/ 2 2 2 1 1 1 / kx x y + 1 y = y 2 , 2 y = ( ) 3/ 2 2 2 1 1 1 / ky x y + 8. Let 1 2 1 1 2 1 2 , , , so 4 2 3 , x x x x x y y y y y x x x y x ′′ = = = = = = = = − + 2 3 2 cos . y y x y y t ′′ = = + Equivalent system: 1 x = x 2 , 2 x = - 4 x 1 + 2 y 1 - 3 x 2 1 y = y 2 , 2 y = 3 x 1 - y 1 - 2 y 2 + cos t 9. Let 1 2 1 1 2 1 1 2 1 , , , , , , so x x x x x y y y y y z z z z z = = = = = = = = = 2 3 2 , x x x y z ′′ = = + 2 2 4 , 5 . y y x y z z z x y z ′′ ′′ = = + = = Equivalent system: 1 x = x 2 , 2 x = 3 x 1 - y 1 + 2 z 1 1 y = y 2 , 2 y = x 1 + y 1 - 4 z 1 1 z = z 2 , 2 z = 5 x 1 - y 1 - z 1 10. Let 1 2 1 1 2 1 2 , , , so (1 ), x x x x x y y y y y x x x y ′′ = = = = = = = = 2 (1 ). y y y x ′′ = = Equivalent system: 1 x = x 2 , 2 x = x 1 (1 - y 1 ) 1 y = y 2 2 y = y 1 (1 - x 1 ) 11. The computation x'' = y' = - x yields the single linear second-order equation x'' + x = 0 with characteristic equation r 2 + 1 = 0 and general solution x ( t ) = A cos t + B sin t . Then the original first equation y = x' gives
Section 7.1 335 y ( t ) = B cos t - A sin t . The figure on the left below shows a direction field and typical solution curves (obviously circles?) for the given system. 12. The computation x'' = y' = x yields the single linear second-order equation x'' - x = 0 with characteristic equation r 2 - 1 = 0 and general solution x ( t ) = A e t + B e - t . Then the original first equation y = x' gives y ( t ) = A e t - B e - t . The figure on the right above shows a direction field and some typical solution curves of this system. It appears that the typical solution curve is a branch of a hyperbola.

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