Engr 213 midterm 1b sol 2010

Concordia University February 9, 2010

Applied Ordinary Differential Equations
ENGR 213 - Section F
Prof. Alina Stancu

Exam I (B)

(1) (10 points) Solve the initial value problem

dy
(x2 + 4) + 2xy = x, y(0) = 1.
dx
Solution: An integrating factor is
2x
dx 2 +4)
µ(x) = e x2 +4 = eln(x = x2 + 4.
Multiplying both sides of the ODE with this integrating factor leads to
(x2 + 4) y = x.
Integrating both sides with respect to x, we obtain
x2 x2 /2 + C
(x2 + 4) y = +C ⇒ y = , C = constant.
2 x2 + 4
We use y(0) = 1 to find the constant C as y(0) = C/4 = 1 ⇒ C = 4, thus the
solution to the IVP is
x2 /2 + 4 x2 + 8
y(x) = 2 or y(x) = 2 .
x +4 2x + 8

(2) (10 points) Solve the given homogeneous equation by the appropriate substitution

dy 3x2 + 2y 2
= .
dx 4xy

You may leave the solution in implicit form.
Solution: We use the standard substitution for homogeneous ODEs of first order,
dy
namely u = y/x or y = ux and, thus, dx = x du + u.
dx
Therefore the ODE becomes
du 3 + 2u2 du 3 − 2u2
x +u= ⇒ x = .
dx 4u dx 4u
4u 1
The latter is a separable ODE 3−2u2 du = x dx. Integrating, we obtain − ln |3−2u2 | =
ln |x| + C, where C is an arbitrary constant.
Hence we obtain the solution y(x), in implicit form, of the original ODE
1

2

− ln |3 − 2(y/x)2 | = ln |x| + C, C = constant.

(3) (10 points) Solve the exact ordinary differential equation

√ x y
x+ dx + dy = 0,
x2 + y 2 + 1 x2 + y 2 + 1

leaving the solution in implicit form.
Solution: We seek a function φ(x, y) such that
∂φ √ x ∂φ y
= x+ , = .
∂x x 2 + y2 + 1 ∂y x2 + y2 + 1
Integrating the first equation with respect to x, we obtain
φ(x, y) = 2 x3/2 + x2 + y 2 + 1 + c(y). Consequently,
3
∂φ y
= + c (y) ⇒ c (y) = 0 ⇒ c(y) = C, C = constant.
∂y x2 + y 2 + 1
It follows that the solution to the exact ODE (given in implicit form) is
2 3/2 ˜ ˜
x + x2 + y 2 + 1 = C, C = constant.
3

(4) (10 points) A pitcher of buttermilk initially at 25◦ C is to be cooled by setting it on
the front porch, where the temperature is 5◦ C. Suppose that the temperature of the
buttermilk has dropped to 15◦ C after 20 minutes. When will it be at 10◦ C?
Solution: Newton’s model of cooling tells us that the ODE describing the tempera-
ture of the cake T (t) is
dT
= −K(T − 5), T (0) = 25, T (20) = 15, K = a constant to be determined.
dt
We solve first for the general solution of the ODE (which we treat as separable):
1
dT = −K dt ⇒ ln |T − 5| = −Kt + C, C = constant.
T −5
Using T (0) = 25 and T (20) = 15, we obtain C = ln 20 and ln 10 = −20K + ln 20 ⇒
20K = ln 2 ⇒ K = (ln 2)/20.
Thus
t/20
1
T = 5 + e−Kt+C = 5 + e−t(ln 2)/20+ln 20 = 5 + 20
2
and

t/20
1
10 = 5 + 20 ⇒ t = 20 ln(5/20)/ ln(1/2) = 20 ln(4)/ ln 2 = 40 min.
2

Engr 213 midterm 1b sol 2010

More Related Content

What's hot (20)

Similar to Engr 213 midterm 1b sol 2010 (20)

Engr 213 midterm 1b sol 2010