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Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 332
Question 332 of 490
Easy
MCQ
The solution of
d
2
y
d
x
2
=
e
−
2
x
\dfrac{d^2y}{dx^2}=e^{-2x}
d
x
2
d
2
y
=
e
−
2
x
is:
A
e
−
2
x
4
\dfrac{e^{-2x}}{4}
4
e
−
2
x
B
e
−
2
x
4
+
c
x
+
d
\dfrac{e^{-2x}}{4}+cx+d
4
e
−
2
x
+
c
x
+
d
C
1
4
e
−
2
x
+
c
x
2
+
d
\dfrac14 e^{-2x}+cx^2+d
4
1
e
−
2
x
+
c
x
2
+
d
D
1
4
e
−
4
x
+
c
x
+
d
\dfrac14 e^{-4x}+cx+d
4
1
e
−
4
x
+
c
x
+
d
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Solutions · 1
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Official solution
Integrating twice:
d
y
d
x
=
−
e
−
2
x
2
+
c
1
\dfrac{dy}{dx}=-\dfrac{e^{-2x}}{2}+c_1
d
x
d
y
=
−
2
e
−
2
x
+
c
1
;
y
=
e
−
2
x
4
+
c
1
x
+
c
2
=
e
−
2
x
4
+
c
x
+
d
y=\dfrac{e^{-2x}}{4}+c_1x+c_2 = \dfrac{e^{-2x}}{4}+cx+d
y
=
4
e
−
2
x
+
c
1
x
+
c
2
=
4
e
−
2
x
+
c
x
+
d
.
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