JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Differential Equations

y = y (x)

is the solution of the differential equation

\frac{5 + e ^{x}}{2 + y} \cdot \frac{d y}{d x} + e^{x} = 0

satisfying

y (0) = 1

, then a value of

y (lo g_{e} 13)

is:

Answer: (C)

Solution

Separate the variables:

\frac{d y}{2 + y} = - \frac{e ^{x}}{5 + e ^{x}} d x

. Integrating,

ln (2 + y) = - ln (5 + e^{x}) + ln C

, so

2 + y = \frac{C}{5 + e ^{x}}

. The condition

y (0) = 1

gives

3 = \frac{C}{6}

, hence

C = 18

and

2 + y = \frac{18}{5 + e ^{x}}

. At

x = lo g_{e} 13

2 + y = \frac{18}{5 + 13} = 1

, so

y = - 1

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