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Mathematics
Mathematics
Problem
If
y
=
e
(
k
+
1
)
x
y = e^{(k+1)x}
y
=
e
(
k
+
1
)
x
is a solution of differential equation
d
2
y
d
x
2
−
4
d
y
d
x
+
4
y
=
0
\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0
d
x
2
d
2
y
−
4
d
x
d
y
+
4
y
=
0
, then
k
=
k =
k
=
(A) -1 (B) 0 (C) 1 (D) 2
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