Higher Engineering Mathematics, Sixth Edition

Linear first order differential equations 457

48.2 Procedure to solve differential

equations of the form

dy

dx

+Py=Q

(i) Rearrange the differential equation into the form

dy

dx

+Py=Q,wherePandQare functions ofx.

(ii) Determine

∫

Pdx.

(iii) Determine the integrating factor e

∫

Pdx.

(iv) Substitute e

∫

Pdxinto equation (3).

(v) Integrate the right hand side of equation (3) to

give the general solution of the differential equa-

tion. Given boundary conditions, the particular

solution may be determined.

48.3 Worked problems on linear first

order differential equations

Problem 1. Solve

1

x

dy

dx

+ 4 y=2giventhe

boundary conditionsx=0wheny=4.

Using the above procedure:

(i) Rearranging gives

dy

dx

+ 4 xy= 2 x, which is of the

form

dy

dx

+Py=QwhereP= 4 xandQ= 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

( 2 x)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 equation

dy

dx

+ 1 =−

y

x

is given byy=

3 −x^2

2 x

,given

x=1wheny=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.afunctionofx).

(ii)

∫

Pdx=

∫

1

x

dx=lnx.

(iii) Integrating factore

∫

Pdx=elnx=x(from the def-

inition 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=0wheny=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=1andQ=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.