(i) y = ax^3, a is arbitrary constant. Find differential equation by eliminating arbitrary constant.
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y=Ax
3
+Bx
2
dx
dy
=3Ax
2
+2Bx
x
1
dx
dy
=3Ax+2B -- Equation (1)
Differentiating again, we get
x
2
−1
dx
dy
+
x
1
dx
2
d
2
y
=3A --Equation (2)
Putting values of A&B in Equation 1 & 2 respectively in y=Ax
3
+Bx
2
, we get
As A=
3
1
(
x
1
dx
2
d
2
y
−
x
2
1
dx
dy
)
2B=
x
1
dx
dy
−3Ax
2B=
x
1
dx
dy
−
dx
2
d
2
y
+
x
1
dx
dy
=(
x
2
dx
dy
−
dx
2
d
2
y
)
B=(
x
1
dx
dy
−
2
1
dx
2
d
2
y
)
Now, Ax
3
+Bx
2
=
3
1
x
2
dx
2
d
2
y
−
3
1
x
dx
dy
+x
dx
dy
−
2
1
x
2
dx
2
d
2
y
Ax
3
+Bx
2
=y=
6
−1
x
2
dx
2
d
2
y
+
3
2
x
dx
dy
6y=x(4
dx
dy
)−x
2
dx
2
d
2
y
or
x
2
dx
2
d
2
y
−4x
dx
dy
+6y=0
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