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BITSAT 2024
Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Differential Equations
BITSAT 2024 · Paper 2 · Mathematics · Differential Equations
If (d y)/(d x)-y _e 2=2^ x ( x-1) _e 2 , then y
If
d
y
d
x
−
y
log
e
2
=
2
sin
x
(
cos
x
−
1
)
log
e
2
\frac{d y}{d x}-y \log _e 2=2^{\sin x}(\cos x-1) \log _e 2
d
x
d
y
−
y
lo
g
e
2
=
2
s
i
n
x
(
cos
x
−
1
)
lo
g
e
2
, then
y
y
y
Options
A
2
sin
x
+
c
2
x
2^{\sin x}+c 2^x
2
s
i
n
x
+
c
2
x
B
2
cos
x
+
c
2
x
2^{\cos x}+c 2^x
2
c
o
s
x
+
c
2
x
C
2
sin
x
+
c
2
−
x
2^{\sin x}+c 2^{-x}
2
s
i
n
x
+
c
2
−
x
D
2
cos
x
+
c
2
−
x
2^{\cos x}+c 2^{-x}
2
c
o
s
x
+
c
2
−
x
Correct answer
A.
2
sin
x
+
c
2
x
2^{\sin x}+c 2^x
2
s
i
n
x
+
c
2
x
Step-by-step solution
d
y
d
x
−
y
log
e
2
=
2
sin
x
(
cos
x
−
1
)
log
e
2
\frac{d y}{d x}-y \log _e 2=2^{\sin x}(\cos x-1) \log _e 2
d
x
d
y
−
y
lo
g
e
2
=
2
s
i
n
x
(
cos
x
−
1
)
lo
g
e
2
This is linear differential equation
I.F.
=
e
−
log
e
2
∫
d
x
=
e
−
x
log
e
2
=
2
−
x
\text{ I.F. }=e^{-\log _e 2 \int d x}=e^{-x \log _e 2}=2^{-x}
I.F.
=
e
−
l
o
g
e
2
∫
d
x
=
e
−
x
l
o
g
e
2
=
2
−
x
then general Solution is
y
2
−
x
=
∫
2
−
x
2
sin
x
(
cos
x
−
1
)
log
e
2
d
x
+
c
y 2^{-x}=\int 2^{-x} 2^{\sin x}(\cos x-1) \log _e 2 d x+c
y
2
−
x
=
∫
2
−
x
2
s
i
n
x
(
cos
x
−
1
)
lo
g
e
2
d
x
+
c
Now let
sin
x
−
x
=
t
⇒
(
cos
x
−
1
)
d
x
=
d
t
\sin x-x=t \Rightarrow(\cos x-1) d x=d t
sin
x
−
x
=
t
⇒
(
cos
x
−
1
)
d
x
=
d
t
∴
y
2
−
x
=
log
e
2
∫
2
t
d
t
+
c
∴
y
2
−
x
=
2
t
+
c
∴
y
=
2
x
+
t
+
c
2
x
∴
y
=
2
sin
x
+
c
2
x
\begin{aligned} & \therefore y 2^{-x}=\log _e 2 \int 2^t d t+c \\ & \therefore y 2^{-x}=2^t+c \\ & \therefore y=2^{x+t}+c 2^x \\ & \therefore y=2^{\sin x}+c 2^x \end{aligned}
∴
y
2
−
x
=
lo
g
e
2
∫
2
t
d
t
+
c
∴
y
2
−
x
=
2
t
+
c
∴
y
=
2
x
+
t
+
c
2
x
∴
y
=
2
s
i
n
x
+
c
2
x
$$
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