Question

Let y(x) be a solution of the differential equation $(1+e^x)y^{\prime }+y{e}^x=1.Ify\left(0\right)=2,$ then which of the following statement (s) is/are true ?

• A. y (-4) = 1
• B. y(-2) = 0
• C. y(x) has a critical point in the interval (-1, 0)
• D. y(x) has no critical point in the interval (-1, 0)

Answer 1

The correct option is C

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

Here $(1+e^x)y^{\prime }+y{e}^x=1\Rightarrow \frac{dy}{dx}+e^x.\frac{dy}{dx}+y{e}^x=1\Rightarrow dy+e^{x}dy+y{e}^{x}dx=dx\Rightarrow dy+d\left(e^{x}y\right)=dxOnintegratingbothsides,wegety+e^{x}y=x+CGiven,y\left(0\right)=2\Rightarrow 2+e^0.2=0+C\Rightarrow C=4\therefore y(1+e^x)=x+4\Rightarrow y=\frac{x+4}{1+e^x}Nowatx=-4,y=\frac{-4+4}{1+e^{-4}}=0\therefore y(-4)=0Forcriticalpoints,\frac{dy}{dx}=0i.e.\frac{dy}{dx}=\frac{(1+e^x).1-e^x(x+4)}{(1+e^x{)}^2}=0\Rightarrow e^x(x+3)-1=0Or{e}^{-x}=(x+3)$

Clearly, the intersection point lies between (-1, 0).
$\therefore$ y(x) has a critical point in the interval (-1, 0).