458 Higher Engineering Mathematics
(iv) Substituting in equation (3) gives:
ye
x
=
e
x
(x) dx
(4)
(v)
e x (x) dx is determined using integration by
parts (see Chapter 43).
xe
x dx = xe
x
− e
x
+ c
Hence from equation (4): ye x = xe x − e x + c,
which is the general solution.
When x = 0, y = 2 thus 2e 0 = 0 − e 0 + c, from
which, c = 3.
Hence the particular solution is:
ye
x
= xe
x
− e
x
+ 3 or y = x − 1 + 3e
−x
Now try the following exercise
Exercise 182 Further problems on linear
first order differential equations
Solve the following differential equations.
1. x
dy
dx
= 3 − y
y = 3 +
c
x
2.
dy
dx
= x(1 − 2y)
y =
1
2 + ce −x 2
3. t
dy
dt
−5t =−y
y =
5t
2
+
c
t
4. x
dy
dx
+ 1
= x 3 − 2y, given x = 1 when
y = 3
y =
x
3
5
−
x
3
+
47
15x 2
5.
1
x
dy
dx
+ y = 1
y = 1 +ce −x 2 /2
6.
dy
dx
+ x = 2y
y =
x
2
+
1
4
+ ce 2x
48.4 Further worked problems on
linear first order differential
equations
Problem 4. Solve the differential equation
dy
dθ
= sec θ + y tan θ given the boundary conditions
y = 1 when θ = 0.
Using the procedure of Section 48.2:
(i) Rearranging gives
dy
dθ
− (tan θ)y = sec θ, which is
of the form
dy
dθ
+ P y = Q where P =−tan θ and
Q = sec θ.
(ii)
P dx =
− tan θdθ =−ln(sec θ)
= ln(sec θ) −1 = ln(cos θ).
(iii) Integrating
factor
e
P dθ = e ln(cosθ) = cos θ
(from the definition of a logarithm).
(iv) Substituting in equation (3) gives:
y cos θ =
cos θ(sec θ)dθ
i.e. y cos θ =
dθ
(v) Integrating gives: y cos θ = θ + c, which is the
general solution. When θ = 0, y = 1, thus
1 cos0 = 0 +c, from which, c = 1.
Hence the particular solution is:
y cos θ = θ + 1 or y = (θ + 1) sec θ
Problem 5.
(a) Find the general solution of the equation
(x − 2)
d y
dx
+
3(x − 1)
(x + 1)
y = 1
(b) Given the boundary conditions that y = 5 when
x =−1, find the particular solution of the
equation given in (a).
(a) Using the procedure of Section 48.2:
(i) Rearranging gives:
dy
dx
+
3(x − 1)
(x + 1)(x − 2)
y =
1
(x − 2)
which is of the form
dy
dx
+ P y = Q, where P =
3(x − 1)
(x + 1)(x − 2)
and Q =
1
(x − 2)
(ii)
P dx =
3(x − 1)
(x + 1)(x − 2)
dx, which is
integrated using partial fractions.
(iv) Substituting in equation (3) gives:
ye
x
=
e
x
(x) dx
(4)
(v)
e x (x) dx is determined using integration by
parts (see Chapter 43).
xe
x dx = xe
x
− e
x
+ c
Hence from equation (4): ye x = xe x − e x + c,
which is the general solution.
When x = 0, y = 2 thus 2e 0 = 0 − e 0 + c, from
which, c = 3.
Hence the particular solution is:
ye
x
= xe
x
− e
x
+ 3 or y = x − 1 + 3e
−x
Now try the following exercise
Exercise 182 Further problems on linear
first order differential equations
Solve the following differential equations.
1. x
dy
dx
= 3 − y
y = 3 +
c
x
2.
dy
dx
= x(1 − 2y)
y =
1
2 + ce −x 2
3. t
dy
dt
−5t =−y
y =
5t
2
+
c
t
4. x
dy
dx
+ 1
= x 3 − 2y, given x = 1 when
y = 3
y =
x
3
5
−
x
3
+
47
15x 2
5.
1
x
dy
dx
+ y = 1
y = 1 +ce −x 2 /2
6.
dy
dx
+ x = 2y
y =
x
2
+
1
4
+ ce 2x
48.4 Further worked problems on
linear first order differential
equations
Problem 4. Solve the differential equation
dy
dθ
= sec θ + y tan θ given the boundary conditions
y = 1 when θ = 0.
Using the procedure of Section 48.2:
(i) Rearranging gives
dy
dθ
− (tan θ)y = sec θ, which is
of the form
dy
dθ
+ P y = Q where P =−tan θ and
Q = sec θ.
(ii)
P dx =
− tan θdθ =−ln(sec θ)
= ln(sec θ) −1 = ln(cos θ).
(iii) Integrating
factor
e
P dθ = e ln(cosθ) = cos θ
(from the definition of a logarithm).
(iv) Substituting in equation (3) gives:
y cos θ =
cos θ(sec θ)dθ
i.e. y cos θ =
dθ
(v) Integrating gives: y cos θ = θ + c, which is the
general solution. When θ = 0, y = 1, thus
1 cos0 = 0 +c, from which, c = 1.
Hence the particular solution is:
y cos θ = θ + 1 or y = (θ + 1) sec θ
Problem 5.
(a) Find the general solution of the equation
(x − 2)
d y
dx
+
3(x − 1)
(x + 1)
y = 1
(b) Given the boundary conditions that y = 5 when
x =−1, find the particular solution of the
equation given in (a).
(a) Using the procedure of Section 48.2:
(i) Rearranging gives:
dy
dx
+
3(x − 1)
(x + 1)(x − 2)
y =
1
(x − 2)
which is of the form
dy
dx
+ P y = Q, where P =
3(x − 1)
(x + 1)(x − 2)
and Q =
1
(x − 2)
(ii)
P dx =
3(x − 1)
(x + 1)(x − 2)
dx, which is
integrated using partial fractions.