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Joint Entrance Examination
Mathematics
Differential Equations
Question 148
Question 148 of 490
Medium
MCQ
Let
y
=
y
(
t
)
y=y(t)
y
=
y
(
t
)
solve
d
y
d
t
+
α
y
=
γ
e
−
β
t
\dfrac{dy}{dt}+\alpha y=\gamma e^{-\beta t}
d
t
d
y
+
α
y
=
γ
e
−
β
t
,
α
,
β
,
γ
>
0
\alpha,\beta,\gamma>0
α
,
β
,
γ
>
0
. Then
lim
t
→
∞
y
(
t
)
\lim_{t\to\infty}y(t)
lim
t
→
∞
y
(
t
)
:
A
is 0
B
does not exist
C
is 1
D
is -1
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Official solution
I.F.
=
e
α
t
=e^{\alpha t}
=
e
α
t
.
y
e
α
t
=
γ
α
−
β
e
(
α
−
β
)
t
+
c
⟹
y
=
γ
α
−
β
e
−
β
t
+
c
e
−
α
t
ye^{\alpha t} = \dfrac{\gamma}{\alpha-\beta}e^{(\alpha-\beta)t}+c \implies y = \dfrac{\gamma}{\alpha-\beta}e^{-\beta t}+ce^{-\alpha t}
y
e
α
t
=
α
−
β
γ
e
(
α
−
β
)
t
+
c
⟹
y
=
α
−
β
γ
e
−
β
t
+
c
e
−
α
t
. Since
α
,
β
>
0
\alpha,\beta>0
α
,
β
>
0
, both terms
→
0
\to0
→
0
as
t
→
∞
t\to\infty
t
→
∞
, so
lim
y
(
t
)
=
0
\lim y(t)=0
lim
y
(
t
)
=
0
.
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