458 Higher Engineering Mathematics
(iv) Substituting in equation (3) gives:
yex=
∫
ex(x)dx (4)
(v)
∫
ex(x)dx is determined using integration by
parts (see Chapter 43).
∫
xexdx=xex−ex+c
Hence from equation (4): yex=xex−ex+c,
which is the general solution.
Whenx=0, y=2 thus 2e^0 = 0 −e^0 +c, from
which,c=3.
Hence the particular solution is:
yex=xex−ex+ 3 ory=x− 1 +3e−x
Now try the following exercise
Exercise 182 Further problemson linear
first order differential equations
Solve the following differential equations.
- x
dy
dx
= 3 −y
[
y= 3 +
c
x
]
2.
dy
dx
=x( 1 − 2 y)
[
y=^12 +ce−x
2 ]
- t
dy
dt
− 5 t=−y
[
y=
5 t
2
+
c
t
]
- x
(
dy
dx
+ 1
)
=x^3 − 2 y,givenx=1when
y= 3
[
y=
x^3
5
−
x
3
+
47
15 x^2
]
5.
1
x
dy
dx
+y= 1
[
y= 1 +ce−x
(^2) / 2 ]
6.
dy
dx
+x= 2 y
[
y=
x
2
- 1
4
+ce^2 x
]
48.4 Further worked problems on
linear first order differential
equations
Problem 4. Solve the differential equation
dy
dθ
=secθ+ytanθgiven the boundary conditions
y=1whenθ=0.
Using the procedure of Section 48.2:
(i) Rearranging gives
dy
dθ
−(tanθ)y=secθ,whichis
of the form
dy
dθ
+Py=QwhereP=−tanθand
Q=secθ.
(ii)
∫
Pdx=
∫
−tanθdθ=−ln(secθ)
=ln(secθ)−^1 =ln(cosθ).
(iii) Integrating factor e
∫
Pdθ=eln(cosθ)=cosθ
(from the definition of a logarithm).
(iv) Substituting in equation (3) gives:
ycosθ=
∫
cosθ(secθ)dθ
i.e. ycosθ=
∫
dθ
(v) Integrating gives:ycosθ=θ+c, which is the
general solution. When θ=0, y=1, thus
1cos0= 0 +c, from which,c=1.
Hence the particular solution is:
ycosθ=θ+ 1 ory=(θ+1)secθ
Problem 5.
(a) Find the general solution of the equation
(x− 2 )
dy
dx
+
3 (x− 1 )
(x+ 1 )
y= 1
(b) Given the boundary conditions thaty=5when
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
+Py=Q,whereP=
3 (x− 1 )
(x+ 1 )(x− 2 )
andQ=
1
(x− 2 )
(ii)
∫
Pdx=
∫
3 (x− 1 )
(x+ 1 )(x− 2 )
dx,whichis
integrated using partial fractions.