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Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 152
Question 152 of 490
Hard
MCQ
The solution of
d
y
d
x
=
−
x
2
+
3
y
2
3
x
2
+
y
2
\dfrac{dy}{dx}=-\dfrac{x^2+3y^2}{3x^2+y^2}
d
x
d
y
=
−
3
x
2
+
y
2
x
2
+
3
y
2
,
y
(
1
)
=
0
y(1)=0
y
(
1
)
=
0
is:
A
log
e
∣
x
+
y
−
x
y
(
x
+
y
)
2
∣
=
0
\log_e\left|x+y-\dfrac{xy}{(x+y)^2}\right|=0
lo
g
e
x
+
y
−
(
x
+
y
)
2
x
y
=
0
B
log
e
∣
x
+
y
+
x
y
(
x
+
y
)
2
∣
=
0
\log_e\left|x+y+\dfrac{xy}{(x+y)^2}\right|=0
lo
g
e
x
+
y
+
(
x
+
y
)
2
x
y
=
0
C
log
e
∣
x
+
y
+
2
x
y
(
x
+
y
)
2
∣
=
0
\log_e\left|x+y+\dfrac{2xy}{(x+y)^2}\right|=0
lo
g
e
x
+
y
+
(
x
+
y
)
2
2
x
y
=
0
D
log
e
∣
x
+
y
−
2
x
y
(
x
+
y
)
2
∣
=
0
\log_e\left|x+y-\dfrac{2xy}{(x+y)^2}\right|=0
lo
g
e
x
+
y
−
(
x
+
y
)
2
2
x
y
=
0
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Solutions · 1
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Official solution
With
y
=
v
x
y=vx
y
=
v
x
: eventually
2
x
y
(
x
+
y
)
2
+
ln
(
x
+
y
)
=
c
\dfrac{2xy}{(x+y)^2}+\ln(x+y)=c
(
x
+
y
)
2
2
x
y
+
ln
(
x
+
y
)
=
c
; using
x
=
1
,
y
=
0
x=1,y=0
x
=
1
,
y
=
0
:
c
=
0
c=0
c
=
0
, giving option C.
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