MATH 258 - Introduction to Differential Equations Review Exercises

Çankaya University
Department of Mathematics
MATH 258 - Introduction to Differential Equations
Review Exercises
01.12.2016
1) Find a general solution to given differential equations.
a) y 00 − y 0 − 2y = 0.
Ans. y(x) = c1 e2x + c2 e−x
b) 4w00 + 20w0 + 25w = 0.
Ans. w(x) = c1 e−5x/2 + c2 xe−5x/2
c) z 00 − 4z 0 + 7z = 0
√ √ Ans. z(x) = c1 e2x cos 3x + c2 e2x sin 3x
d) y 000 + y 00 + 3y 0 − 5y = 0
Ans. y(x) = c1 ex + c2 e−x cos (2x) + c3 e−x sin (2x)
2) Solve the given IVP’s
a) y 00 − 4y 0 − 5y = 0, y(−1) = 3 and y 0 (−3) = 9.
Ans. y(x) = 2e5(x+1) + e−(x+1)
b) y 00 + 2y 0 + y = 0, y(0) = 1 and y 0 (0) = −3.
Ans. y(x) = e−x − 2xe−x
c) y 000 − 4y 00 + 7y 0 − 6y = 0, y(0) = 1, y 0 (0) = 0 and y 00 (0) = 0.
√ √
Ans. y(x) = e2x − 2ex sin 2x
3) Find a particular solution to given differential equations.
a) 4y 00 + 11y 0 − 3y = −2xe−3x .
8
x
Ans. yp (x) =
+
xe−3x
13 169
b) y 00 + 2y 0 + 4y = 111e2x cos (3x) .
Ans. yp (x) = e2x cos (3x) + 6e2x sin (3x)
c) y 000 − y 00 + y = sin x
Ans. yp (x) =
1
2
cos x + sin x
5
5
d) y 000 + y 00 − 2y = xex
Ans. yp (x) =
4
1 2 x
x e − xex
10
25
4) Find the solution to the given IVP’s.
a) y 00 − y = sin x − e2x , y(0) = 1 and y 0 (0) = −1.
1
3
7
1
Ans. y(x) = − sin x − e2x + ex + e−x
2
3
4
12
b) y 0 − y = 1, y(0) = 0.
Ans. y(x) = ex − 1
c) y 00 + y = 2e−x , y(0) = 0 and y 0 (0) = 0.
Ans. y(x) = e−x − cos x + sin x
5) Find a general solution to given differential equations.
a) y 00 + y = sec x.
Ans. y(x) = cos x ln | cos x| + x sin x + c1 cos x + c2 sin x
b) y 00 + 4y 0 + 4y = e−2x ln x.
Ans. y(x) =
(2 ln x − 3) x2 e−2x
+ c1 e−2x + c2 xe−2x
4
c) y 00 + y = tan2 x.
Ans. y(x) = sin x ln | sec x + tan x| − 2 + c1 cos x + c2 sin x
6) Find a general solution to given differential equations for x > 0.
a) x2 y 00 + 2xy 0 − 6y = 0.
Ans. y(x) = c1 x−3 + c2 x2
b) y 00 +
6 0
4
y + 2 y = 0.
x
x
Ans. y(x) = c1 x−1 + c2 x−4
c) x2 y 00 + xy 0 + 7y = − tan (3 ln x).
Ans. y(x) = c1 cos (3 ln x) + c2 sin (3 ln x) +
1
cos (3 ln x) ln | sec (3 ln x) + tan (3 ln x) |
9
3
d) x y − xy + y = x 1 +
.
ln x
2 00
0
1
Ans. y(x) = c1 x + c2 x ln x + x (ln x)2 + 3x ln x (ln | ln x|)
2
7) Given that y1 (x) = x is a solution to
y 00 −
1 0
1
y + 2 y = 0,
x
x
use the reduction of oreder procedure to determine a second linearly independent solution for
x > 0.
Ans. y2 (x) = x ln x
8) Given that y1 (x) = x + 1 is a solution to
xy 00 − (x + 1)y 0 + y = x2 ,
find the general solution.
Ans. y = c1 ex + c2 (1 + x) − x2
9) Given that y1 (x) = e−5x is a solution to
xy 00 + (5x − 1)y 0 − 5y = x2 e−5x ,
find the general solution.
Ans. y = c1 (5x − 1) + c2 e−5x −
10) Given that y1 (x) =
x2 e−5x
10
1
is a solution to
x
x2 y 00 − 2xy 0 − 4y = 0,
find the general solution.
Ans. y = c1 x4 + c2
1
x