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# Chapter 17 - INSTRUCTORS SOLUTIONS MANUAL SECTION 17.1(PAGE...

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 17.1 (PAGE 902) CHAPTER 17. ORDINARY DIFFEREN- TIAL EQUATIONS Section 17.1 Classifying Differential Equations (page 902) 1. dy dx = 5 y : 1st order, linear, homogeneous. 2. d 2 y dx 2 + x = y : 2nd order, linear, nonhomogeneous. 3. y dy dx = x : 1st order, nonlinear. 4. y + xy = x sin x : 3rd order, linear, nonhomogeneous. 5. y + x sin x y = y : 2nd order, linear, homogeneous. 6. y + 4 y 3 y = 2 y 2 : 2nd order, nonlinear. 7. d 3 y dt 3 + t dy dt + t 2 y = t 3 : 3rd order, linear, nonhomogeneous. 8. cos x dx dt + x sin t = 0: 1st order, nonlinear, homogeneous. 9. y ( 4 ) + e x y = x 3 y : 4th order, linear, homogeneous. 10. x 2 y + e x y = 1 y : 2nd order, nonlinear. 11. If y = cos x , then y + y = − cos x + cos x = 0. If y = sin x , then y + y = − sin x + sin x = 0. Thus y = cos x and y = sin x are both solutions of y + y = 0. This DE is linear and homogeneous, so any function of the form y = A cos x + B sin x , where A and B are constants, is a solution also. There- fore sin x cos x is a solution ( A = − 1, B = 1), and sin ( x + 3 ) = sin 3 cos x + cos 3 sin x is a solution, but sin 2 x is not since it cannot be repre- sented in the form A cos x + B sin x . 12. If y = e x , then y y = e x e x = 0; if y = e x , then y y = e x e x = 0. Thus e x and e x are both solutions of y y = 0. Since y y = 0 is linear and homogeneous, any function of the form y = Ae x + Be x is also a solution. Thus cosh x = 1 2 ( e x + e x ) is a solu- tion, but neither cos x nor x e is a solution. 13. Given that y 1 = cos ( kx ) is a solution of y + k 2 y = 0, we suspect that y 2 = sin ( kx ) is also a solution. This is easily verified since y 2 + k 2 y 2 = − k 2 sin ( kx ) + k 2 sin ( kx ) = 0 . Since the DE is linear and homogeneous, y = Ay 1 + By 2 = A cos ( kx ) + B sin ( kx ) is a solution for any constants A and B . It will satisfy 3 = y (π/ k ) = A cos (π) + B sin (π) = − A 3 = y (π/ k ) = − Ak sin (π) + Bk cos (π) = − Bk , provided A = − 3 and B = − 3 / k . The required solution is y = − 3 cos ( kx ) 3 k sin ( kx ). 14. Given that y 1 = e kx is a solution of y k 2 y = 0, we suspect that y 2 = e kx is also a solution. This is easily verified since y 2 k 2 y 2 = k 2 e kx k 2 e kx = 0 . Since the DE is linear and homogeneous, y = Ay 1 + By 2 = Ae kx + Be kx is a solution for any constants A and B . It will satisfy 0 = y ( 1 ) = Ae k + Be k 2 = y ( 1 ) = Ake k Bke k , provided A = e k / k and B = − e k / k . The required solution is y = 1 k e k ( x 1 ) 1 k e k ( x 1 ) . 15. By Exercise 11, y = A cos x + B sin x is a solution of y + y = 0 for any choice of the constants A and B . This solution will satisfy 0 = y (π/ 2 ) 2 y ( 0 ) = B 2 A , 3 = y (π/ 4 ) = A 2 + B 2 , provided A = 2 and B = 2 2. The required solution is y = 2 cos x + 2 2 sin x . 627

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SECTION 17.1 (PAGE 902) R. A. ADAMS: CALCULUS 16. y = e rx is a solution of the equation y y 2 y = 0 if r 2 e rx re rx 2 e rx = 0, that is, if r 2 r 2 = 0. This quadratic has two roots, r = 2, and r = − 1.
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Chapter 17 - INSTRUCTORS SOLUTIONS MANUAL SECTION 17.1(PAGE...

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