Higher Engineering Mathematics

SOLUTION OF FIRST ORDER DIFFERENTIAL EQUATIONS BY SEPARATION OF VARIABLES 445

I

Since 5

dy

dx

+ 2 x=3 then

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 and
x=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=2 whent= 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), wherewand

xare 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 178 Further problems on equa-

tions of the form

dy

dx

=f(x).

In Problems 1 to 5, solve the differential

equations.

1.

dy

dx

=cos 4x− 2 x

[

y=

sin 4x

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=2 whenx=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

]

5.

1

ex

+ 2 =x− 3

dy

dx

,giveny=1 whenx=0.

[

y=

1

6

(

x^2 − 4 x+

2

ex

+ 4

)]

6. The gradient of a curve is given by:
   dy
   dx

+

x^2

2

= 3 x