JEE Main 2023 · Mathematics · Differential Equations · hard

Let y(x) satisfy dy/dx=y+y^2e^x with y(0)=1. Then y(-ln 2) equals

Single correct option

Let

y(x)

satisfy

\dfrac{dy}{dx}=y+y^2e^{x}

with

y(0)=1

. Then

y(-\ln 2)

equals

Show answer and solution

Answer: (C) $\dfrac{4}{11}$

This is a Bernoulli equation. Divide by

y^2

\frac{y'}{y^2}=\frac1y+e^x

. With

u=\frac1y

u'=-\frac{y'}{y^2}

, so

u'+u=-e^x

. The integrating factor is

e^x

(ue^x)'=-e^{2x}

, hence

ue^x=-\frac{e^{2x}}{2}+C

. From

y(0)=1

1=-\frac12+C

, so

C=\frac32

and

u=\frac{3e^{-x}-e^{x}}{2}

. At

x=-\ln2

e^{-x}=2

and

e^{x}=\frac12

, so

u=\frac{6-1/2}{2}=\frac{11}{4}

and

y=\frac{4}{11}

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