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Joint Entrance Examination
Mathematics
Differential Equations
Question 253
Question 253 of 490
Medium
MCQ
The differential equation of the family of curves
x
2
=
4
b
(
y
+
b
)
x^2=4b(y+b)
x
2
=
4
b
(
y
+
b
)
,
b
∈
R
b\in\mathbb{R}
b
∈
R
, is:
A
x
(
y
′
)
2
=
x
+
2
y
y
′
x(y')^2=x+2yy'
x
(
y
′
)
2
=
x
+
2
y
y
′
B
x
(
y
′
)
2
=
2
y
y
′
−
x
x(y')^2=2yy'-x
x
(
y
′
)
2
=
2
y
y
′
−
x
C
x
y
′
′
=
y
′
xy''=y'
x
y
′′
=
y
′
D
x
(
y
′
)
2
=
x
−
2
y
y
′
x(y')^2=x-2yy'
x
(
y
′
)
2
=
x
−
2
y
y
′
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Official solution
2
x
=
4
b
y
′
⇒
b
=
x
2
y
′
2x=4by' \Rightarrow b=\dfrac{x}{2y'}
2
x
=
4
b
y
′
⇒
b
=
2
y
′
x
. Substituting:
x
2
=
2
x
y
′
y
+
x
2
(
y
′
)
2
⟹
x
(
y
′
)
2
=
2
y
y
′
+
x
x^2 = \dfrac{2x}{y'}y+\dfrac{x^2}{(y')^2} \implies x(y')^2 = 2yy'+x
x
2
=
y
′
2
x
y
+
(
y
′
)
2
x
2
⟹
x
(
y
′
)
2
=
2
y
y
′
+
x
.
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