MATH2341 Exam 1 Answers Feb 27, 2025
Name:
1. Find the general explicit solution for the following.
(a) (3 points)
4x3 − 24x2 y 2 dx + − 16x3 y + 2 cos(2y) dy = 0.
Solution: This is an exact equation,
Z
ϕ(x, y) = 4x3 − 24x2 y 2 dx = x4 − 8x3 y 2 + g(y)
=⇒ ϕy = −16x3 y + g ′ (y) = −16x3 y + 2 cos(2y) =⇒ g(y) = sin(2y)
The solution is
x4 − 8x3 y 2 + sin(2y) = C.
(b) (4 points)
2 √
y ′ − y = 4x5 y.
x
√
Solution: This is Bernoulli, divide by y to get
√
−1/2 ′ 2 y
y y − = 4x5 .
x
√ y −1/2 ′
Let u = y =⇒ u′ = 2
y =⇒ y −1/2 y ′ = 2u′ .
2u u
2u′ − = 4x5 =⇒ u′ − = 2x5 .
x x
The I.F is I(x) = x−1 , multiply this on both sides and do backward prod rule
to get
′ 2x5 2x6 2x6 2
x−1 u = 2x4 =⇒ x−1 u = +C =⇒ u = +Cx =⇒ y = +Cx .
5 5 5
Page 1 of 5
, MATH2341 Exam 1 Contd.
2. (4 points) Find the general solution for the following second order DE:
y ′′ − 3y ′ = 12x
Solution: Here
yh = C1 + C2 e3x .
Since r = 0 is a root of the CE, we guess
yp = Ax2 + Bx =⇒ yp′ = 2Ax + B =⇒ yp′′ = 2A.
Plug into the DE and we get,
4
2A − 6Ax − 3B = 12x =⇒ A = −2, B = − .
3
The general solution is
4x
y = C1 + C2 e3x − 2x2 − .
3
3. A spring mass dashpot system has m = 1, c = 6 and k = 10. There is no external force
on the system.
(a) (3 points) If x(0) = −2 and x′ (0) = 3, find the equation of motion x(t) for this
system.
Solution: The DE is x′′ + 6x′ + 10x = 0 and the solution is
−3t
x(t) = e C1 cos(t) + C2 sin(t)
Plug in the initial values to get
x(t) = e−3t − 2 cos(t) − 3 sin(t)
(b) (2 points) Write the solution in amplitude-phase form.
Solution:
−3t
√ −3 √
x(t) = e 13 cos(t − arctan( ) + π = e−3t 13 cos(t − 4.122)
−2
Page 2 of 5
Name:
1. Find the general explicit solution for the following.
(a) (3 points)
4x3 − 24x2 y 2 dx + − 16x3 y + 2 cos(2y) dy = 0.
Solution: This is an exact equation,
Z
ϕ(x, y) = 4x3 − 24x2 y 2 dx = x4 − 8x3 y 2 + g(y)
=⇒ ϕy = −16x3 y + g ′ (y) = −16x3 y + 2 cos(2y) =⇒ g(y) = sin(2y)
The solution is
x4 − 8x3 y 2 + sin(2y) = C.
(b) (4 points)
2 √
y ′ − y = 4x5 y.
x
√
Solution: This is Bernoulli, divide by y to get
√
−1/2 ′ 2 y
y y − = 4x5 .
x
√ y −1/2 ′
Let u = y =⇒ u′ = 2
y =⇒ y −1/2 y ′ = 2u′ .
2u u
2u′ − = 4x5 =⇒ u′ − = 2x5 .
x x
The I.F is I(x) = x−1 , multiply this on both sides and do backward prod rule
to get
′ 2x5 2x6 2x6 2
x−1 u = 2x4 =⇒ x−1 u = +C =⇒ u = +Cx =⇒ y = +Cx .
5 5 5
Page 1 of 5
, MATH2341 Exam 1 Contd.
2. (4 points) Find the general solution for the following second order DE:
y ′′ − 3y ′ = 12x
Solution: Here
yh = C1 + C2 e3x .
Since r = 0 is a root of the CE, we guess
yp = Ax2 + Bx =⇒ yp′ = 2Ax + B =⇒ yp′′ = 2A.
Plug into the DE and we get,
4
2A − 6Ax − 3B = 12x =⇒ A = −2, B = − .
3
The general solution is
4x
y = C1 + C2 e3x − 2x2 − .
3
3. A spring mass dashpot system has m = 1, c = 6 and k = 10. There is no external force
on the system.
(a) (3 points) If x(0) = −2 and x′ (0) = 3, find the equation of motion x(t) for this
system.
Solution: The DE is x′′ + 6x′ + 10x = 0 and the solution is
−3t
x(t) = e C1 cos(t) + C2 sin(t)
Plug in the initial values to get
x(t) = e−3t − 2 cos(t) − 3 sin(t)
(b) (2 points) Write the solution in amplitude-phase form.
Solution:
−3t
√ −3 √
x(t) = e 13 cos(t − arctan( ) + π = e−3t 13 cos(t − 4.122)
−2
Page 2 of 5
