Question

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

• A. $y(-4)=0$
• 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

Answer: (A), (C)

$\left(1+e^x\right)y\prime +y{e}^x=1$
$\Rightarrow d\left(\left(1+e^x\right)y\right)=1$
$\Rightarrow y\left(1+e^x\right)=x+C$
$\because y\left(0\right)=2$
$\therefore 2\times 2=C\Rightarrow C=4$
$\therefore y\left(1+e^x\right)=x+4$
$\therefore y\left(-4\right)=0,y\left(-2\right)=\frac{2}{1+e^{-2}}\ne 0$
For critical point y′ = 0
$\Rightarrow y{e}^x=1$
Now let g$\left(x\right)=y{e}^x-1=\frac{e^x\left(x+4\right)}{1+e^x}-1$
$g\left(-1\right)=\frac{3e^{-1}}{1+e^{-1}}-1=\frac{3}{e+1}-1<0$
$g\left(0\right)=2-1>0$
So there exists one value of x in $\left(-1,0\right)$ for which g$\left(x\right)=0\Rightarrow y\prime =0$
$\Rightarrow$ there exist a critical point of $y\left(x\right)$ in $\left(-1,0\right)$