# BÀI TẬP ĐÀN HỒI ỨNG DỤNG

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1. Second-order; linear. 2. Third-order; nonlinear because of (dy/dx) 4. 3. The differential equation is first-order. Writing it in the khung 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 size u(dv/du) + (1 + u)v = ue u we see that it is linear in v. However, writing it in the size (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 2y + y = −e −x/2 + e −x/2 = 0. 12. From y = 6 5 − 6 5 e −20t we obtain dy/dt = 24e −20t , so that dy dt + 20y = 24e −20t + đôi mươi 6 5 − 6 5 e −20t = 24. 13. From y = e 3x cos 2x we obtain y = 3e 3x cos 2x − 2e 3x sin 2x and y = 5e 3x cos 2x − 12e 3x sin 2x, so that y − 6y + 13y = 0. 14. From y = − cos x ln(sec x + chảy x) we obtain y = −1 + sin x ln(sec x + chảy x) và y = tan x + cos x ln(sec x + chảy x). Then y + y = rã x. 15. Writing ln(2X − 1) − ln(X − 1) = t and differentiating implicitly we obtain 2 2X − 1 dX dt − 1 X − 1 dX dt = 1 2 2X − 1 − 1 X − 1 dX dt = 1 2X − 2 − 2X + 1 (2X − 1)(X − 1) dX dt = 1 dX dt = −(2X − 1)(X − 1) = (X − 1)(1 − 2X).

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Exponentiating both sides of the implicit solution we obtain 2X − 1 X − 1 = e t =⇒ 2X − 1 = xe cộ 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 = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid. 