Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A $y(-4)=0$
C $y\left(x\right)$ has a critical point in the interval $(-1,0)$

$(1+e^x)y^{\prime }+y{e}^x=1\frac{dy}{dx}+\frac{e^x}{1+e^x}y=\frac{1}{1+e^x}\text{I.F.}=e^{\int \frac{e^x}{1+e^x}dx}=1+e^x\therefore y(1+e^x)=\int \frac{1+e^x}{1+e^x}dx+C\Rightarrow y=\frac{C+x}{1+e^x}$

Given $y\left(0\right)=2\Rightarrow C=4$
$\therefore y=\frac{4+x}{1+e^x}\Rightarrow y(-4)=\frac{4-4}{1+e^x}=0$
So, option (1) is correct.

For critical points, $\frac{dy}{dx}=0$
$y=\frac{4+x}{1+e^x}\frac{dy}{dx}=\frac{(1+e^x)-(x+4)e^x}{(1+e^x{)}^2}=0\Rightarrow 1-3e^x-x{e}^x=0\Rightarrow 3+x=e^{-x}$

Let $f\left(x\right)=e^{-x}-x-3$
Then $f\left(0\right)<0\text{and}f(-1)>0$
$f$ is continuous and changes sign from negative to positive in the interval $(-1,0).$
$\therefore f\left(x\right)$ has a root in $(-1,0).$
So, we can say that $y\left(x\right)$ has a critical point in $(-1,0).$