1. Midterm Exam Emat 213
February 2006
Instructor: Dr. Marco Bertola
Time allowed: 1h15min.
Material allowed: calculators.
Recommendations: use only blue or black ink
Solve four problems: the top score is 40 points
Clearly indicate which problems you wish marked
[10 points] Problem 1.
Find the general solution of the following separable ODE
2
dy y2 + 1 3
= x2 .
dx y
Solution
y
2 dy = x2 dx
(y 2 + 1) 3
3 2 1 x3
(y + 1) 3 = +C
2 3
[10 points] Problem 2.
Determine which of the following ODE’s is exact and then solve it (in implicit form).
(a) (x2 + y 2 + ey+x )dx − ey+x + y 3 dy = 0
(b) 2x − 4e2x+3y dx + 3y 2 − 6e2x+3y dy = 0
Solution (just answer)
The second is exact (because the cross derivatives match). The solution reads (implicitly)
x2 − 2e2x+3y + y 3 = C
[10 points] Problem 3.
(i) Which of the following two first order equations is linear? Explain why the other is not linear
dy 2 e−x
(a) + 1+ y= 2
dx x x
dy
(b) − x = cos(y)
dx
(ii) Find the general solution of the linear equation that you have found above.
Solution
The second is nonlinear because the unknown function y is precomposed with cos.

2. The solution of the linear one is
C −x e−x
y= e +
x2 x
[10 points] Problem 4.
Perform the substitution u = y −1 in the following Bernoulli ODE so as to obtain a new linear ODE in the
dependent variable u and find the general solution in term of y
dy
x2 − xy = y 2 .
dx
Solution
1 u′
y= ; y′ = −
u u2
u′ x 1
−x2 2 − = 2
u u u
′ 1 1
u + u=− 2
x x
C
uc =
x
1 1 ln |x|
up = − x 2 dx = −
x x x
C ln |x|
u= −
x x
1 x
y= =
C
x − ln |x| x
C − ln |x|
[10 points] Problem 5.
A cake is removed from an oven at a temperature of 150 degrees (Celsius) and left to cool down on a countertop
in a room at a temperature of 20 degrees. After 2 minutes the temperature of the cake is 120 degrees. After
how many minutes (approximately) the cake’s temperature will have dropped to 40 degrees?
Solution
The equation of Newton says
T (t) = Ce−kt + Tenv
We have
Tenv = 20 ; T (0) = 150 =⇒ C + 20 = 150 ; C = 130;
T (2) = 120 =⇒ 130e−2k = 100
1 10
k = − ln
2 13
The solution is obtained by solving for t the following equation
T (t) = 40

3. 1 10
130 exp ln t + 20 = 40
2 13
2
ln 13
t=2 10 ≃ 14.25 mins
ln 13
[10 points] Problem 6.
Solve the following equation by using a substitution of the form u = Ax + By + C
y ′ = (−2x + y)2 − 7
Solution
u = −2x + y
y = u + 2x
y ′ = u′ + 2
′
u + 2 = u2 − 7
u′ = u2 − 9
du
= dx = x + C
u2 −9
1 u−3
ln =x+C
6 u+3
(by partial fraction expansion ....)
u−3 ˜
= Ce6x
u+3
y − 2x − 3 ˜
= Ce6x
y − 2x + 3