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Instructor: Math 10560, Worksheet: Separable Differential Equations February 29, 2016
- Please show all of your work for all questions both MC and PC
- work without using a calculator.
- Multiple choice questions should take about 4 minutes to complete.
- Partial credit questions should take about 8 minutes to complete.
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
- (a) (b) (c) (d) (e)
- (a) (b) (c) (d) (e) ........................................................................................................................
- (a) (b) (c) (d) (e)
- (a) (b) (c) (d) (e) ........................................................................................................................
- (a) (b) (c) (d) (e)
Please do NOT write in this box. Multiple Choice
Total
Instructor:
Multiple Choice
1.(6 pts) The solution to the initial value problem
(x^2 + 1) dydx = y, y(0) = 2
satisfies the implicit equation
(a) ln |y| = tan−^1 x + ln 2 (b) ln |y| = tan−^1 x
(c) ln |y| = (ln 2) tan−^1 x (d) y^2 = 2 tan−^1 x
(e) y^2 = 2 tan−^1 x + 4
2.(6 pts) Find the solution of the differential equation: dy dx
x + 1 ey^
with initial condition y(0) = 2. First we separate the variables:
eydy = (x + 1)dx.
Integrating both sides yields ey^ = x^2 /2 + x + C. To solve for y we take the log of both sides:
y = ln(x^2 /2 + x + C).
Finally, we use the initial condition y(0) = 2 to solve for C.
2 = ln(C) =⇒ C = e^2.
Thus y = ln(x^2 /2 + x + e^2 ). Note that ln(x^2 /2 + x + e^2 ) = ln|x^2 /2 + x + e^2 | since x^2 /2 + x + e^2 > 0.
(a) y = 2 + e([x
(^2) /2] + x)
(b) y = 2e([x
(^2) /2] + x)
(c) y = ln
x^2 2
∣ + 2^ (d)^ y^ = ln
x^2 2
(e) y = 2 + ln | ln | x^2 2
Instructor:
and ln |y| = x^2 + x + C Applying the exponential function to both sides, we get
|y| = ex (^2) +x+C = eC^ ex (^2) +x = Kex (^2) +x
we can discard the absolute values sign and absorb it in the constant.
y = Kex (^2) +x
From the initial value, we get 2 = Ke^0 = K Hence our solution is y = 2ex (^2) +x
(a) y = 2ex(x+1)^ (b) y = 2 ln(y) x^2 + ln 2
(c) y = 2ex(2x+1)^ (d) y = ex 2
(e) y = ex(x+1)^ + 1
Instructor:
5.(6 pts) Solve the initial value problem dy dx = y^2 y(0) = − 1. This ODE is separable. Hence we have ∫ dy y^2
dx,
so −
y = x + C ⇔ y =
x + C and −1 = y(0) =
0 + C
, hence C = 1.
(a) y =
x (b) y = 0 (c) y = −x^2
(d) y =
x + 1 (e) y =
x + 1
Instructor:
- (13 pts.) Find the family of orthogonal trajectories to the family of curves given by
y = k
x.
Instructor:
- (13 pts.) Find the family of orthogonal trajectories to the family of curves given by
y = kx^2.
dy dx
= 2kx
For the family of curves given above
y = kx^2 giving k = y x^2 Thus this family of curves satisfy the differential equation dy dx
y x^2 x = 2 y x
Now using the fact that the product of the derivatives of two orthogonal curves meeting at a point must equal −1, we get that the orthogonal trajectories satisfy the differential equation dy dx
−x 2 y
Separating the variables, we get 2 ydy = −xdx and
2
ydy = −
xdx, or y^2 = −x^2 2
+ C.
Hence our family of orthogonal trajectories is a family of curves of the form
y^2 + x^2 2
= C,
a family of ellipses.
Instructor: ANSWERS Math 10560, Worksheet: Separable Differential Equations February 29, 2016
- Please show all of your work for all questions both MC and PC
- work without using a calculator.
- Multiple choice questions should take about 4 minutes to complete.
- Partial credit questions should take about 8 minutes to complete.
PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!
- (•) (b) (c) (d) (e)
- (a) (b) (c) (•) (e) ........................................................................................................................
- (a) (b) (•) (d) (e)
- (•) (b) (c) (d) (e) ........................................................................................................................
- (a) (b) (c) (•) (e)
Please do NOT write in this box. Multiple Choice
Total
