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 differential equation is first-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 differential equation is first-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. Fourth-order; linear 6. Second-order; nonlinear because of cos( r + u ) 7. Second-order; nonlinear because of 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. From y = e x/ 2 we obtain y = 1 2 e x/ 2 . Then 2 y + y = e x/ 2 + e x/ 2 = 0. 12. From y = 6 5 6 5 e 20 t we obtain dy/dt = 24 e 20 t , so that dy dt + 20 y = 24 e 20 t + 20 6 5 6 5 e 20 t = 24 . 13. From y = e 3 x cos2 x we obtain y = 3 e 3 x cos2 x 2 e 3 x sin2 x and y = 5 e 3 x cos2 x 12 e 3 x sin2 x , so that y 6 y + 13 y = 0. 14. From y = cos x ln(sec x + tan x ) we obtain y = 1 + sin x ln(sec x + tan x ) and y = tan x + cos x ln(sec x + tan x ). Then y + y = tan x . 15. Writing ln(2 X 1) ln( X 1) = t and differentiating implicitly we obtain 2 2 X 1 dX dt 1 X 1 dX dt = 1 2 2 X 1 1 X 1 dX dt = 1 2 X 2 2 X + 1 (2 X 1)( X 1) dX dt = 1 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 1) = ( e t 2) X = X = e t 1 e t 2 . Solving e t 2 = 0 we get t = ln2. Thus, the solution is defined on ( −∞ , ln2) or on (ln2 , ). The graph of the solution defined on ( −∞ , ln2) is dashed, and the graph of the solution defined on (ln2 , ) is solid. 1

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-4 -2 2 4 x -4 -2 2 4 y Exercises 1.1 16. Implicitly differentiating the solution we obtain 2 x 2 dy dx 4 xy + 2 y dy dx = 0 = ⇒ − x 2 dy 2 xy dx + y dy = 0 = 2 xy dx + ( x 2 y ) dy = 0 . 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 defined on ( −∞ , ). The graph of y 1 ( x ) is solid and the graph of y 2 is dashed. 17. Differentiating 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 ] 1 + c 1 e t = P (1 P ) . 18. Differentiating y = e x 2 x 0 e t 2 dt + c 1 e x 2 we obtain y = e x 2 e x 2 2 xe x 2 x 0 e t 2 dt 2 c 1 xe x 2 = 1 2 xe x 2 x 0 e t 2 dt 2 c 1 xe x 2 . Substituting into the differential equation, we have y + 2 xy = 1 2 xe x 2 x 0 e t 2 dt 2 c 1 xe x 2 + 2 xe x 2 x 0 e t 2 dt + 2 c 1 xe x 2 = 1 .
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