Higher Engineering Mathematics, Sixth Edition

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.

1. x

dy

dx

= 3 −y

[

y= 3 +

c

x

]

2.

dy

dx

=x( 1 − 2 y)

[

y=^12 +ce−x

2 ]

3. t

dy

dt

− 5 t=−y

[

y=

5 t

2

+

c

t

]

4. 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.