Ecuaciones Diferenciales

DAMARIS BETHSABE MEDINA NIETO Assignment D1Tarea1Var sep homog due 02/17/2013 at 05:32pm EST

1. (1 pt) Find the equation of the solution to through the point (x, y) = (1, 3).

dy = x6 y dx

20132 Ec Dif ByF AL

4. (1 pt) Solve the initial value problem y 0 =

x2 − y2 with xy

y(2) = 1.

Correct Answers: • y = 3 / (eˆ(1/7)) eˆ(xˆ7 / 7)

Correct Answers: • y = sqrt(xˆ2/2-4/(xˆ2))

2. (1 pt) Sometimes a change of variable can be used to convert a differential equation y 0 = f (t, y) into a separable equation. One common change of variable technique is as follows. (1) Consider a differential equation of the form y 0 = f (αt + βy + γ), where α, β, and γ are constants. Use the change of variable z = αt + βy + γ to rewrite the differential equation as a separable equation of the form z 0 = g(z).

5. (1 pt) Solve the initial value problem y 0 = (x + y − 3)2 with y(0) = 0. (1) To solve this, we should use the substitution u= u0 = Enter derivatives using prime notation (e.g., you would dy enter y 0 for dx ).

Solve the initial value problem

(2) After the substitution from the previous part, we obtain the following linear differential equation in x, u, u 0 .

y 0 = (t + y)2 − 1, y(3) = 5.

(a) g(z) =

(3) The solution to the original initial value problem is described by the following equation in x, y.

(b) y(t) =

Correct Answers:

Correct Answers: • zˆ2 • (-1) / (t - 3 - 1/8) - t

• • • •

dy x = . dx 49y (1) Find an implicit solution and put your answer in the following form: = constant.

3. (1 pt) Solve the differential equation

x+y-3 1+y’ u’ = uˆ2+1 y = tan(x + arctan(-3)) - x + 3

6. (1 pt) Solve the initial value problem 2yy 0 + 2 = y2 + 2x with y(0) = 8. (1) To solve this, we should use the substitution u=

(2) Find the equation of the solution through the point (x, y) = (−7, 1).

With this substitution, y=

(3) Find the equation of the solution through the point (x, y) = (0, −5). Your answer should be of the form y = f (x).

y0 = Enter derivatives using prime notation (e.g., you would dy enter y 0 for dx ).

Correct Answers: • yˆ2-(x/7)ˆ2 • x+7*y = 0 • y = -sqrt( (x/7)ˆ2 + 25 )

(2) After the substitution from the previous part, we obtain the following linear differential equation in x, u, u 0 . 1

(3) The solution to the original initial value problem is described by the following equation in x, y.

10. (1 pt) Find a non-constant solution to (x 0 )2 + x2 = 9 using your knowledge of derivatives from basic calculus.

Correct Answers: • • • • •

x(t) =

yˆ2 sqrt(u) (1/2) uˆ(-1/2) u’ u’-u = 2*x-2 y = sqrt( 64 eˆx - 2 x)

Correct Answers: • 3*cos(t)

11. (1 pt) Find a solution to

7. (1 pt) Solve y 0 = (y − 1)(y + 1) if y(2) = 0.

Correct Answers: • y = k * eˆ(xˆ2/2 + 3 x) - 7

y(x) =

12. (1 pt)√ Solve the initial value problem yy 0 + x = with y(1) = 3. (1) To solve this, we should use the substitution u=

Correct Answers: • 2 / (1 - eˆ(2x-4)) - 1

8. (1 pt) Find a solution to

dy = xy + 7x + 3y + 21. dx

dA = −9A if A(0) = 3. dt

p

x 2 + y2

u0 =

A(t) = Correct Answers:

Enter derivatives using prime notation (e.g., you would dy ). enter y 0 for dx

• 3*eˆ(-9*t)

(2) After the substitution from the previous part, we obtain the following linear differential equation in x, u, u 0 .

9. (1 pt) Suppose y(t) = 7e−5t is a solution of the initial value problem y 0 + ky = 0, y(0) = y0 . What are the constants k and y0 ? k=

(3) The solution to the original initial value problem is described by the following equation in x, y.

y0 =

Correct Answers: • xˆ2+yˆ2 • 2*x+2*y*y’ • u’ = 2*sqrt(u) • y = sqrt(2*x+1)

Correct Answers: • 5 • 7

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