Higher Engineering Mathematics

456 DIFFERENTIAL EQUATIONS

equation. Given boundary conditions, the par-

ticular solution may be determined.

48.3 Worked problems on linear first

order differential equations

Problem 1. Solve

1

x

dy

dx

+ 4 y=2 given the

boundary conditionsx=0 wheny=4.

Using the above procedure:

(i) Rearranging gives

dy

dx

+ 4 xy= 2 x, which is

of the form

dy

dx

+Py=QwhereP= 4 xand

Q= 2 x.

(ii)

∫

Pdx=

∫

4 xdx= 2 x^2.

(iii) Integrating factor e

∫

Pdx=e 2 x^2.

(iv) Substituting into equation (3) gives:

ye^2 x

2

=

∫

e^2 x

2

(2x)dx

(v) Hence the general solution is:

ye^2 x

2

=^12 e^2 x

2

+c,

by using the substitutionu= 2 x^2 Whenx=0,

y=4, thus 4e^0 =^12 e^0 +c, from which,c=^72.

Hence the particular solution is

ye^2 x

2

=^12 e^2 x

2

+^72

ory=^12 +^72 e−^2 x

2

ory=^12

(

1 +7e−^2 x

2 )

Problem 2. Show that the solution of the equa-

tion

dy

dx

+ 1 =−

y

x

is given byy=

3 −x^2

2 x

,given

x=1 wheny=1.

Using the procedure of Section 48.2:

(i) Rearranging gives:

dy

dx

+

(

1

x

)

y=−1, which

is of the form

dy

dx

+Py=Q, whereP=

1

x

and

Q=−1. (Note thatQcan be considered to be

− 1 x^0 , i.e. a function ofx).

(ii)

∫

Pdx=

∫

1

x

dx=lnx.

(iii) Integrating factor e

∫

Pdx=elnx=x(from the

definition of logarithm).

(iv) Substituting into equation (3) gives:

yx=

∫

x(−1) dx

(v) Hence the general solution is:

yx=

−x^2

2

+c

When x=1, y=1, thus 1=

− 1

2

+c, from

which,c=

3

2

Hence the particular solution is:

yx=

−x^2

2

+

3

2

i.e. 2yx= 3 −x^2 andy=

3 −x^2

2 x

Problem 3. Determine the particular solution

of

dy

dx

−x+y=0, given thatx=0 wheny=2.

Using the procedure of Section 48.2:

(i) Rearranging gives

dy

dx

+y=x, which is of the

form

dy

dx

+P,=Q, whereP=1 andQ=x. (In

this casePcan be considered to be 1x^0 , i.e. a

function ofx).

(ii)

∫

Pdx=

∫

1dx=x.

(iii) Integrating factor e

∫

Pdx=ex.

(iv) Substituting in equation (3) gives:

yex=

∫

ex(x)dx (4)

(v)

∫

ex(x)dxis determined using integration by

parts (see Chapter 43).

∫

xexdx=xex−ex+c