Solutions Manual for Differential Equations and Boundary Value Problems Computing and Modeling 5th E
Published on Aug 5, 2018
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1. Second order; linear 2. Third order; nonlinear because of (dy/dx) 4 3. Fourth order; linear 4. Second order; nonlinear because of cos(r + u) 5. Second order; nonlinear because of (dy/dx) 2 or 1 + (dy/dx) 2 6. Second order; nonlinear because of R 2 7. Third order; linear 8. Second order; nonlinear because of ˙ x 2 9. Writing the differential equation 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. 10. Writing the differential equation 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. 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 + 20 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 + 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. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y = 1 + 2(x + 2) −1/2 we have (y − x)y = (y − x)[1 + (2(x + 2) −1/2 ] = y − x + 2(y − x)(x + 2) −1/2 = y − x + 2[x + 4(x + 2) 1/2 − x](x + 2) −1/2 = y − x + 8(x + 2) 1/2 (x + 2) −1/2 = y − x + 8.