Mukuba University
Department of Mathematics
Mat 220 - Analytic Geometry and Calculus
Tutorial Sheet - ODEs
1. Solve the following DEs by separating variables method.
x2
0
(a) y =
y
1
0
(b) y =
xy + y
0
(c) y + 2x = 5xy
0
(d) y = xe2y
0
(e) (x2 + 1)y = (y 2 + 1)
(f) (y + 2)dx + y(x + 4)dy = 0, y(−3) = −1
2. For the following DEs check whether the DE is homogeneous or not.If yes solve the DE.
(a) xy 2 dy = (x3 + y 3 )dx
(b) (x2 + xy)dx − y 2 dy = 0
0
(c) (x − y) + xy = 0
xy + y 2
0
(d) y =
x2
0
(e) (x2 − y 2 )y = xy
1
0
(f) xy = x − y, y(2) =
2
3. For the following DEs check whether the DE is Exact or not.If yes solve the DE.
(a)
(b)
(c)
(d)
(e)
(3x + 2y)dx + (2x + y)dy = 0
(2xy + 1)dx + (x2 + 4y)dy = 0
0
(3x2 + 1) + (3y 2 + 2y)y = 0
(ysec2 x + secxtanx)dx + (tanx + 2y)dy = 0
0
(ycosx + 2xey ) + (sinx + x2 ey − 1)y = 0
π
(f) (θ2 + 1)cosrdr + 2θsinrdθ = 0, r(0) =
2
4. Solve the following first order linear DEs by integrating factor method.
3y
0
(a) y +
= 6x2
x
0
(b) y + 3y = 3x2 e−3x
0
(c) y + 3y = x
0
(d) xy + 4y = x3 − x
dy
2x
4
(e)
+
y=
2
dx 1 + x
(1 + x2 )2
0
(f) y + 3xy = 8x, y(0) = 1
5. Solve the following Bernoulli DEs.
y
0
(a) y − = xy 2
x
y
0
(b) y + = ex y 4
3
dy
(c) x + y = xy 3
dx
dy
(d) x + 2y = x7 y 3
dx
dy
√
(e) x3
+ xy = 12x y
dx
π
dy
− ycotx = y 3 cosecx, y( ) = 1
(f)
dx
2
6. Solve the following second order linear DEs.
00
0
(a) y + 6y + 9y = 0
d2 y
dy
(b)
− 2 + 10y = 0
2
dx
dx
00
0
(c) y − 5y + 6y = 0
00
0
(d) 4y − 5y − 6y = 0
d2 y
dy
(e)
+
6
+ 10y = 0
dx2
dx
00
0
0
(f) y − y − 2y = 0, y(0) = 2, y (0) = 7
7. Solve the following second order linear DEs, using the method of undetermined coefficients.
00
0
y + y − 2y = 3x
00
0
y + 4y = ex
00
0
y + y − 2y = sinx
00
0
y − 5y + 6y = 2x + 3
00
0
y + y − 2y = 2x2 − 1
d2 x
dx
dx
(f)
− 3 + 2x = e−2t , t = 0, x = 0 and
=0
2
dt
dt
dt
8. Solve the following second order linear DEs,using the method of variation of parameters.
(a)
(b)
(c)
(d)
(e)
(a)
(b)
(c)
(d)
(e)
(f)
00
0
y − 5y + 4y = e3x
00
0
y − 3y − 4y = 2
00
0
y − 9y = 6x
00
0
y − 5y + 4y = 2cosx
00
0
3y + y = 2y + x2 − 1
00
0
0
y − y − 2y = e3x , y(0) = 1, y (0) = 2
£££££ End £££££
Page 2