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JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 40
Total: 124
Let the tangent at any point
P
on a curve passing through the points
(
1
,
1
)
and
(
1
10
,
100
)
, intersect positive
x
axis and
y
-axis at the points
A
and
B
respectively. If
P
A
:
P
B
=
1
:
k
and
y
=
y
(
x
)
is the solution of the differential equation
e
dy
dx
=
kx
+
k
2
,
y
(
0
)
=
k
, then
4
y
then
4
y
(
1
)
−
5
log
e
3
is equal to _________.
[10-Apr-2023 shift 2]
Your Answer:
Validate
Solution:
Y
−
y
=
d
y
d
x
(
X
−
x
)
Y
=
0
X
=
−
y
dx
d
y
+
x
k
α
+
0
k
+
1
=
x
,
α
=
k
+
1
k
x
k
+
1
k
x
=
−
y
dx
dy
+
x
x
+
x
k
=
−
y
dx
dy
+
x
x
dy
dx
+
ky
=
0
d
y
d
x
+
k
x
y
=
0
y
⋅
x
k
=
C
C
=
1
100
⋅
(
1
10
)
k
=
1
K
=
2
d
y
d
x
=
ln
(
2
x
+
1
)
y
=
(
2
x
+
1
)
2
(
ln
(
2
x
+
1
)
−
1
)
+
c
2
=
1
2
(
0
−
1
)
+
C
C
=
2
+
1
2
=
5
2
y
(
1
)
=
3
2
(
ℓ
ln
3
−
1
)
+
5
2
=
3
2
ln
3
+
1
4
y
(
1
)
=
6
ln
3
+
4
4
y
(
1
)
−
5
ln
3
=
4
+
ln
3
© examsnet.com
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