Answers of the exam of April 13, 2011
Differential equations (wi2029LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y + 2y + 5y = (t ) cos(t); y(0) = 2, y (0) = 1.
Note: Lcfw_(t c)f (t) = ecs f (c).
Answer:
The tra

Answers of the exam of April 16, 2013
Differential equations (wi2029LR)
1. Question:
Use the Laplace transform to solve the following integro-differential equation
Zt
0
y (t) + y(t) y( ) sin(t )d = sin t , y(0) = 1.
0
Answer:
Note that the third term is a

Answers of the exam of April 18, 2012
Differential equations (wi2029LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y + 4y = u (t) cos(t ) , y(0) = 2, y (0) = 1.
Answer:
The transformed equation is
s2 Y (s) sy(0) y

Answers of the exam of November 1, 2013
Differential equations (wi2180LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y + 4y = (t 10) + 1 , y(0) = 1, y (0) = 0.
Answer:
The transformed equation is
1
s2 Y (s) sy(0)

Answers of the exam of January 26, 2012
Differential equations (WI2029LR)
1. Question:
Use the Laplace transform to solve the following integro-differential equation:
Zt
2
y (t) = t + y(t ) cos d , y(0) = 1 .
0
Answer:
First note that
2
y (t) = t +
Zt
y(t

Answers of the exam of January 23, 2015
Differential equations (wi2180LR)
1. Question:
Consider the integral equation for the function (t)
Zt
(t)
e(t) () d = cos 2t.
0
Use the Laplace transform to solve this integral equation.
Answer:
Rt
One should recog

Answers of the exam of January 24, 2013
Differential equations (wi2029LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y 00 + 2y 0 + 5y = 1 u1 (t); y(0) = 2, y 0 (0) = 2.
Answer:
The transformed equation is
s2 Y (s)

Answers of the exam of October 31, 2014
Differential equations (wi2180LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y 00 + 4y = f (t) , y(0) = 0, y 0 (0) = 1,
where f (t) is defined as
(
f (t) =
0
e(t)
if 0 t <

Faculty EEMCS
Division Applied Mathematics
Mekelweg 4, 2628 CD DELFT
Differential Equations, WI2180LR
Tuesday 3 November 2015: EXAMPLE (short answer part)
Naam:
Grade :
Studienummer:
Your answers should be written in the frames.
Only your final answer is

Answers of the exam of April 14, 2010
Differential equations (wi2029LR)
1. Question:
Use the Laplace transform to solve the following differential equation:
y + 2y = sin(t) + (t ) , y(0) = 1 .
Answer:
The transformed equation is
sY (s) y(0) + 2Y (s) =
1
+

Answers of the exam of January 21, 2011
Differential equations (WI2029LR)
1. Question:
Use the Laplace transform to solve the following initial value problem:
y 2y + 2y = u2 (t)e(t2) ,
y(0) = 0, y (0) = 1.
Answer:
The transformed equation is
s2 Y (s) sy(0