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 12 1 2 and , so 7 3 . x x xxx xx t ′′ == = = = + Equivalent system: 1 x = x 2 , 2 x = - 7 x 1 - 3 x 2 + t 2 2. Let 1 3 2 4 3 ,, , , s o = = = = = (4) 4 36 c o s 3 . 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 () 22 1 2 and , so 1 / . x x x x x t x t x t  = = =  Equivalent system: 1 x = x 2 , t 2 2 x = (1 - t 2 ) x 1 - tx 2 4. Let 23 1 3 2 3 , s o 5 3 2 l n / x xx x x x x x x x x t x tx 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 3 2 3 , s o c o s . x x x = = = = = + Equivalent system: 1 x = x 2 , 2 x = x 3 , 3 x = x 2 2 + cos x 1 6. Let 1 1 2 1 2 2 , s o 5 4 ,4 5 . x xx x x y yy y y x yy 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 22 3 / 2 12 1 1 2 1 2 ,, , s o / ( ) , xx xxx yy yyy k x xy ′′ == = = = = + 3 / 2 2 /( ) . k y + Equivalent system: 1 x = x 2 , 2 x = () 3/2 11 1 / kx x y −+ 1 y = y 2 , 2 y = 1 / ky x y 8. Let 1 1 2 , s o4 2 3 , x xx x x y yy y y x x x y x = = = = + 2 32 c o s . x 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 1 1 , , , , s o y zz zzz = === = 2 , x y z + 4, 5 . x yz 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 1 2 , s o( 1 ) , x x x x y = = = = 2 (1 ). 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
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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. 13. The computation x'' = - 2 y' = - 4 x yields the single linear second-order equation x'' + 4 x = 0 with characteristic equation r 2 + 4 = 0 and general solution x ( t ) = A cos 2 t + B sin 2 t . Then the original first equation y = - x' /2 gives y ( t ) = - B cos 2 t + A sin 2 t . Finally, the condition x (0) = 1 implies that A = 1, and then the condition y (0) = 0 gives B = 0. Hence the desired particular solution is given by -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 x y -5 -4 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 x
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336 Chapter 7 x ( t ) = cos 2 t , y ( t ) = sin 2 t .
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This note was uploaded on 04/08/2008 for the course MATH 374 taught by Professor Zhu during the Spring '08 term at Western Michigan.

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

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