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Joint Entrance Examination
Mathematics
Differential Equations
Question 213
Question 213 of 490
Medium
MCQ
*(Content of question 151 duplicated the previous problem's solution in the source; retained as extracted.)* The differential equation satisfied by the family of parabolas
y
2
=
4
a
(
x
+
a
)
y^2=4a(x+a)
y
2
=
4
a
(
x
+
a
)
is:
A
y
(
d
y
d
x
)
2
−
2
x
d
y
d
x
−
y
=
0
y\left(\dfrac{dy}{dx}\right)^2-2x\dfrac{dy}{dx}-y=0
y
(
d
x
d
y
)
2
−
2
x
d
x
d
y
−
y
=
0
B
y
(
d
y
d
x
)
2
−
2
x
d
y
d
x
+
y
=
0
y\left(\dfrac{dy}{dx}\right)^2-2x\dfrac{dy}{dx}+y=0
y
(
d
x
d
y
)
2
−
2
x
d
x
d
y
+
y
=
0
C
y
(
d
y
d
x
)
2
+
2
x
d
y
d
x
−
y
=
0
y\left(\dfrac{dy}{dx}\right)^2+2x\dfrac{dy}{dx}-y=0
y
(
d
x
d
y
)
2
+
2
x
d
x
d
y
−
y
=
0
D
y
d
y
d
x
+
2
x
d
y
d
x
−
y
=
0
y\dfrac{dy}{dx}+2x\dfrac{dy}{dx}-y=0
y
d
x
d
y
+
2
x
d
x
d
y
−
y
=
0
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Official solution
Differentiating
y
2
=
4
a
x
+
4
a
2
y^2=4ax+4a^2
y
2
=
4
a
x
+
4
a
2
:
2
y
y
′
=
4
a
⟹
a
=
y
y
′
2
2yy'=4a \implies a=\dfrac{yy'}{2}
2
y
y
′
=
4
a
⟹
a
=
2
y
y
′
. Substituting back: \[y^2 = 2yy'\cdot x + y^2(y')^2 \implies y(y')^2+2xy'-y=0.\]
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