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. Show Problems 1.1.1 - 1.3.16 Problems 1.3.17 - 2.2.14 Problems 2.2.15 - 2.4.14 Problems 2.4.15 - 2.6.20 Problems 2.6.21 - 2.9.12 Problems 2.9.13 - 3.1.8 Problems 3.1.9 - 3.2.48 Problems 3.2.49 - 3.4.18 Problems 3.4.19 - 3.5.39 Problems 3.6.1 - 3.8.4 Problems 3.8.5 - 4.2.22 Problems 4.2.23 - 5.1.12 Problems 5.1.13 - 5.3.24 Problems 5.3.25 - 5.5.12 Problems 5.5.13 - 6.1.28 Problems 6.1.29 - 6.3.28 Problems 6.3.29 - 6.6.4 Problems 6.6.5 - 7.2.20 Problems 7.2.21 - 7.5.16 Problems 7.5.17 - 7.8.4 Problems 7.8.5 - 8.1.27 Problems 8.2.1 - 8.5.9 Problems 8.6.1 - 9.3.8 Problems 9.3.9 - 9.6.8 Problems 9.6.9 - 10.2.4 Problems 10.2.5 - 10.4.24 Problems 10.4.25 - 10.7.4 Problems 10.7.5 - 11.2.8 Problems 11.2.9 - 11.6.13
STEM Jock • About • Privacy PolicyCopyright © 2017-2022, stemjock.com. All rights reserved. How do you solve differential equations with boundary conditions?In the earlier chapters we said that a differential equation was homogeneous if g(x)=0 g ( x ) = 0 for all x . Here we will say that a boundary value problem is homogeneous if in addition to g(x)=0 g ( x ) = 0 we also have y0=0 y 0 = 0 and y1=0 y 1 = 0 (regardless of the boundary conditions we use).
What is the solution of the boundary value problem?In mathematics, in the field of differential equations, a boundary value problem is a differential equation together with a set of additional constraints, called the boundary conditions. A solution to a boundary value problem is a solution to the differential equation which also satisfies the boundary conditions.
What is boundary value problem in differential equations?A Boundary value problem is a system of ordinary differential equations with solution and derivative values specified at more than one point. Most commonly, the solution and derivatives are specified at just two points (the boundaries) defining a two-point boundary value problem.
How many solutions does a boundary value problem have?Corollary 51.2 Any homogeneous boundary-value problem has either no solutions, just the constant solution y = 0 , or an infinite number of solutions.
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