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
]
- 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
]
- 3
dy
dθ
+sinθ=0, giveny=
2
3
whenθ=
π
3
[
y=
1
3
cosθ+
1
2
]