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the general solution of d 2 6d 9 y e 3x x 2 where
Question:
The general solution of
(
D
2
+
6
D
+
9
)
y
=
e
−
3
x
x
2
(D^2+6D+9)y=\frac{e^{-3x}}{x^2}
(
D
2
+
6
D
+
9
)
y
=
x
2
e
−
3
x
, where
D
≡
d
d
x
D\equiv \frac{d}{dx}
D
≡
d
x
d
is
(given that c
1
and c
2
are arbitrary constants)
CUET (PG) - 2023
CUET (PG)
Updated On:
Mar 12, 2025
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
e
−
3
x
2
x
y=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
2
x
e
−
3
x
y
=
(
c
1
+
c
2
x
)
e
3
x
+
e
3
x
2
x
y=(c_1+c_2x)e^{3x}+\frac{e^{3x}}{2x}
y
=
(
c
1
+
c
2
x
)
e
3
x
+
2
x
e
3
x
y
=
(
c
1
+
c
2
x
2
)
e
−
3
x
+
e
−
3
x
2
x
y=(c_1+c_2x^2)e^{-3x}+\frac{e^{-3x}}{2x}
y
=
(
c
1
+
c
2
x
2
)
e
−
3
x
+
2
x
e
−
3
x
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
e
3
x
2
x
y=(c_1+c_2x)e^{-3x}+\frac{e^{3x}}{2x}
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
2
x
e
3
x
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The Correct Option is
A
Solution and Explanation
The correct answer is(A):
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
e
−
3
x
2
x
y=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}
y
=
(
c
1
+
c
2
x
)
e
−
3
x
+
2
x
e
−
3
x
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i
n
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