Let y (x) be a solution of the differential equation (1 + ex)y' + yex = 1. If y (0) = 2, then which of the following statements is(are) true?

Engineering

Mathematics

Linear Differential Equation

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?

y (– 4) = 0

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

y (– 2) = 0

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

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

$y' + (\frac{e^{x}}{1 + e^{x}}) y = \frac{1}{1 + e^{x}}$

$I . F . = e^{\int \frac{e^{x}}{1 + e^{x}} dx} = e^{\ln (1 + e^{x})} = 1 + e^{x} .$

$y (1 + ex) = \int 1 \cdot dx + C$

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

Curve is passing through point (0, 2)

$∴$ C = 4

$∴$ y (1 + e^x) = x + 4 ⇒ y = $\frac{x + 4}{1 + e^{x}}$

y(– 4) = 0 ⇒ (A)

$\frac{dy}{dx} = \frac{(1 + e^{x}) \cdot 1 - (x + 4) e^{x}}{{(1 + e^{x})}^{2}} = 0$

⇒ 1 + e^x – (x + 4)e^x = 0

⇒ e^–x + 1 = x + 4

e^–x = x + 3

e^–x – x – 3 = 0

Let g (x) = e^–x – x – 3

g (0) = 1 – 3 (– ve)

g (– 1) = e¹ + 1 – 3 (+ve)

$∴$ e^–x – x – 3 = 0 has a solution in (–1, 0)