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Q.

The general solution of the differential equation, y'+'(x)ϕ(x)ϕ'(x)=0 where ϕ(x) a known function is :

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a

y=ce+ϕ(x)+ϕ(x)1

b

y=ceϕ(x)+ϕ(x)+1

c

y=ceϕ(x)+ϕ(x)1

d

y=ceϕ(x)ϕ(x)+1

answer is A.

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Detailed Solution

dydx+'(x)=ϕ(x)ϕ'(x)

 I.F =eϕ'(x)dx=eϕ(x)

Hence, yeϕ(x)=eϕ(x)ϕ(x)ϕ'(x)dx=ettdt

Where ϕ(x)=t

=tetet+C=ϕ(x)eϕ(x)eϕ(x)+Cy=Ceϕ(x)+ϕ(x)1

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