Question

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 ?

• 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)

$\left(1 + e^{x}\right)y′ + ye^{x} =1$
$⇒ d\left(\left(1 + e^{x}\right)y\right) = 1$
$⇒ y\left(1 + e^{x}\right) = x + C$
$\because y\left(0\right) = 2$
$∴ 2 × 2 = C ⇒ C = 4$
$∴ y\left(1 + e^{x}\right) = x + 4$
$∴ y\left(-4\right) = 0, y\left(-2\right) = \frac{2}{1+e^{-2}}\ne0$
For critical point y′ = 0
$⇒ ye^{x} = 1$
Now let g$\left(x\right) = ye^{x} - 1 = \frac{e^{x}\left(x + 4\right)}{1+e^{x}}-1$
$g\left(-1\right) = \frac{3e^{-1}}{1+e^{-1}}-1 = \frac{3}{e+1}-1< 0$
$g\left(0\right) = 2-1 > 0$
So there exists one value of x in $\left(-1, 0\right)$ for which g$\left(x\right) = 0 ⇒ y′ = 0$
$⇒$ there exist a critical point of $y\left(x\right)$ in $\left(-1, 0\right)$