Chapter 1

Chapter 1 - 1 Introduction to Differential Equations...

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-4 -2 2 4 t -4 -2 2 4 X 1 Introduction to Differential Equations Exercises 1.1 1. Second-order; linear. 2. Third-order; nonlinear because of ( dy/dx ) 4 . 3. The diFerential equation is ±rst-order. Writing it in the form x ( dy/dx )+ y 2 = 1, we see that it is nonlinear in y because of y 2 . However, writing it in the form ( y 2 1)( dx/dy x = 0, we see that it is linear in x . 4. The diFerential equation is ±rst-order. Writing it in the form u ( dv/du )+(1+ u ) v = ue u we see that it is linear in v . However, writing it in the form ( v + uv ue u )( du/dv u = 0, we see that it is nonlinear in u . 5. ²ourth-order; linear 6. Second-order; nonlinear because of cos( r + u ) 7. Second-order; nonlinear because of p 1+( dy/dx ) 2 8. Second-order; nonlinear because of 1 /R 2 9. Third-order; linear 10. Second-order; nonlinear because of ˙ x 2 11. ²rom y = e x/ 2 we obtain y 0 = 1 2 e x/ 2 . Then 2 y 0 + y = e x/ 2 + e x/ 2 =0. 12. ²rom y = 6 5 6 5 e 20 t we obtain dy/dt =24 e 20 t , so that dy dt +20 y e 20 t µ 6 5 6 5 e 20 t . 13. ²rom y = e 3 x cos2 x we obtain y 0 =3 e 3 x x 2 e 3 x sin2 x and y 0 =5 e 3 x x 12 e 3 x x , so that y 0 6 y 0 +13 y 14. ²rom y = cos x ln(sec x + tan x ) we obtain y 0 = 1 + sin x ln(sec x + tan x ) and y 0 = tan x + cos x ln(sec x + tan x ). Then y 0 + y = tan x . 15. Writing ln(2 X 1) ln( X 1)= t and diFerentiating implicitly we obtain 2 2 X 1 dX dt 1 X 1 dX dt =1 µ 2 2 X 1 1 X 1 dX dt 2 X 2 2 X +1 (2 X 1)( X 1) dX dt dX dt = (2 X 1)( X 1)=( X 1)(1 2 X ) . Exponentiating both sides of the implicit solution we obtain 2 X 1 X 1 = e t = 2 X 1= Xe t e t = ( e t e t 2) X = X = e t 1 e t 2 . Solving e t 2 = 0 we get t = ln2. Thus, the solution is de±ned on ( −∞ , ln2) or on (ln2 , ). The graph of the solution de±ned on ( −∞ , ln2) is dashed, and the graph of the solution de±ned on (ln2 , ) is solid. 1
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-4 -2 2 4 x -4 -2 2 4 y Exercises 1.1 16. Implicitly diferentiating the solution we obtain 2 x 2 dy dx 4 xy +2 y dy dx =0 = ⇒− x 2 dy 2 xy dx + ydy =0 = 2 xy dx +( x 2 y ) dy . Using the quadratic Formula to solve y 2 2 x 2 y 1 = 0 For y , we get y = ( 2 x 2 ± 4 x 4 +4 ) / 2= x 2 ± x 4 + 1. Thus, two explicit solutions are y 1 = x 2 + x 4 + 1 and y 2 = x 2 x 4 + 1. Both solutions are de±ned on ( −∞ , ). The graph oF y 1 ( x ) is solid and the graph oF y 2 is dashed. 17. Diferentiating P = c 1 e t / (1 + c 1 e t ) we obtain dP dt = (1 + c 1 e t ) c 1 e t c 1 e t · c 1 e t (1 + c 1 e t ) 2 = c 1 e t 1+ c 1 e t [(1 + c 1 e t ) c 1 e t ] c 1 e t = P (1 P ) . 18. Diferentiating y = e x 2 Z x 0 e t 2 dt + c 1 e x 2 we obtain y 0 = e x 2 e x 2 2 xe x 2 Z x 0 e t 2 dt 2 c 1 xe x 2 =1 2 xe x 2 Z x 0 e t 2 dt 2 c 1 xe x 2 . Substituting into the diferential equation, we have y 0 xy 2 xe x 2 Z x 0 e t 2 dt 2 c 1 xe x 2 xe x 2 Z x 0 e t 2 dt c 1 xe x 2 .
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This homework help was uploaded on 02/13/2008 for the course ENG 342 taught by Professor Delahanty during the Fall '07 term at TCNJ.

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Chapter 1 - 1 Introduction to Differential Equations...

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