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Question

The solution of d2xdy2−x=k, where k is a non-zero constant, vanishes when y=0 and tends of finite limit as y tends to infinity, is

A
x=k(1+ey)
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B
x=k(ey+ey2)
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C
x=k(ey1)
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D
x=k(ey1)
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Solution

The correct option is B x=k(ey1)
We can write given differential equation as,
(D21)x=k...(i)
where, D=ddy
Its auxiliary equation is m21=0, so that
m=1,1
Hence, CF=C1ey+C2ey
where C1,C2 are arbitrary constants
Now, also PI=1D21k
=k.1D21e0.y
=k.1021e0.y=k
So, solution of eq. (i) is
x=C1ey+C2eyk...(ii)
Given that x=0, when y=0
So, 0=C1+C2k (From (ii))
C1+C2=k....(iii)
Multiplying both sides of eq. (ii) by ey, we get
x.ey=C1+C2e2ykey...(iv)
Given that xm when y,m being a finite quantity.
So, eq (iv) becomes
x×0=C1+C2×0(k×0)
C1=0....(v)
From eqs. (iv) and (v), we get
C1=0 and C2=k
Hence, eq. (ii) becomes
x=keyk=k(ey1) which is the required solution.

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