Higher Engineering Mathematics, Sixth Edition

446 Higher Engineering Mathematics

Integrating both sides gives:

y=

∫ (

2

x

− 4 x^2

)

dx

i.e. y=2lnx−

4

3

x^3 +c,

which is the general solution.

Problem 3. Find the particular solution of the

differential equation 5

dy

dx

+ 2 x=3, given the

boundary conditionsy= 1

2

5

whenx= 2.

Since 5

dy

dx

+ 2 x=3then

dy

dx

=

3 − 2 x

5

=

3

5

−

2 x

5

Hence y=

∫(

3

5

−

2 x

5

)

dx

i.e. y=

3 x

5

−

x^2

5

+c,

which is the general solution.

Substituting the boundary conditionsy= 125 andx= 2

to evaluatecgives:

125 =^65 −^45 +c,from which,c= 1

Hence the particular solution isy=

3 x

5

−

x^2

5

+ 1.

Problem 4. Solve the equation

2 t

(

t−

dθ

dt

)

=5, givenθ=2whent= 1.

Rearranging gives:

t−

dθ

dt

=

5

2 t

and

dθ

dt

=t−

5

2 t

Integrating gives:

θ=

∫ (

t−

5

2 t

)

dt

i.e. θ=

t^2

2

−

5

2

lnt+c,

which is the general solution.

Whenθ=2,t=1, thus 2=^12 −^52 ln 1+cfrom which,

c=^32.

Hence the particular solution is:

θ=

t^2

2

−

5

2

lnt+

3

2

i.e. θ=

1

2

(t^2 −5lnt+3)

Problem 5. The bending momentMof the beam

is given by

dM

dx

=−w(l−x),wherewandxare

constants. DetermineMin terms ofxgiven:

M=^12 wl^2 whenx= 0.

dM

dx

=−w(l−x)=−wl+wx

Integrating with respect toxgives:

M=−wlx+

wx^2

2

+c

which is the general solution.

WhenM=^12 wl^2 ,x=0.

Thus

1

2

wl^2 =−wl( 0 )+

w( 0 )^2

2

+c

from which,c=

1

2

wl^2.

Hence the particular solution is:

M=−wlx+

w(x)^2

2

+

1

2

wl^2

i.e. M=

1

2

w(l^2 −2lx+x^2 )

or M=

1

2

w(l−x)^2

Now try the following exercise

Exercise 177 Further problems on

equations of the form

dy

dx

=f(x).

In Problems 1 to 5, solve the differential

equations.

1.

dy

dx

=cos4x− 2 x

[

y=

sin4x

4

−x^2 +c

]

2. 2x

dy

dx

= 3 −x^3

[

y=

3

2

lnx−

x^3

6

+c

]

3.

dy

dx

+x=3, giveny=2whenx=1.

[

y= 3 x−

x^2

2

−

1

2

]

4. 3

dy

dθ

+sinθ=0, giveny=

2

3

whenθ=

π

3

[

y=

1

3

cosθ+

1

2

]