Let y(x) be a solution of the differential equation (1+e^x)y'+ye^x=1.{If}\;y(0)=2,then which of the following statements is (are) true ? [JEE Adv 2015 P1]

y(x) has a critical point in the interval(-1,0)

y(x) has no point in the interval (-1,0)

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Differential Equations

Section: Mathematics

Question : 14 of 23

Marks: +1, -0

Let

y(x)

(1+e^x)y'+ye^x=1.\text{If}\;y(0)=2

Solution:

\frac{dy}{dx}+\frac{ye^x}{1+e^x}=\frac{1}{e^x+1}

I.F=

\int\limits_{e}\frac{e^x}{1+e^x}dx=e^{\ln(1+e^x)}=1+e^x

\Rightarrow y(1+e^x)=\int 1dx

y(1+e^x)=x+c

y=\frac{x+c}{1+e^x}

y(0)=2\Rightarrow c=1

\Rightarrow y=-4=0

\Rightarrow y'=\frac{(1+e^x)-(x+4)e^x}{(1+e^x)^2}=0

Let

g(x)=\frac{(1+e^x)-(x+4)e^x}{(1+e^x)^2}

g(0)=\frac{2-4}{2^2}<0

g(-1)=\frac{\left(1+\frac{1}{e}\right)-\frac{3}{e}}{\left(1+\frac{1}{e}\right)^2}=\frac{1-\frac{2}{e}}{\left(1+\frac{1}{e}\right)^2}>0

g(0).g(-1).Hence g(x)h has a rot in between

(-1,0)

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