Question

2.14
The solution to the differential equation
y lny + xy' = 0, where y(1) = e, is -
(A) x(lny) = 1
(B) xy (lny) = 1
(C) (lny)2 = 2
(D) lny+ (x^2/2 )=1​

Answer 1

Answer:

1

Step-by-step explanation:

ylny + xy' = 0

ylny + xdy/dx = 0

ylnydx + xdy= 0

ylnydx =- xdy

dy/ylny = -dx /x

dy/ylny = -lnx

let lny = t

1/y = dt/dy

dy/y = dt _____(1)

(1) put in equastion

dt/t = -lnx +c

lnt + lnx -c = 0

ln(lny) + lnx - c = 0

ln (lne ) + ln(1) - c = 0 , (given x= 1 ,and y = e)

hence, c = 0

now, equation , ln(lny)+ lnx = 0

ln ( xlny ) = 0

xlny = e^0 = 1

Hence, option (A) is correct ⛄

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@Rajukumar111

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