JEE Advanced | 2015Differential EquationHard
Question
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 ?
the following statements is(are) true ?
Options
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)
Solution
y′ + exy′ + yex = 1
⇒ dy + d(exy) = dx
⇒ y + exy = x + c
∵ y(0) = 2 ⇒ c = 4
⇒ y =
∴ y(-4) = 0
for critical point
given
∵
= 0 ⇒ ex(x + 3) - 1 = 0
⇒ x + 3 = e-x
y(x) has a critical point in the interval (-1, 0)
⇒ dy + d(exy) = dx
⇒ y + exy = x + c
∵ y(0) = 2 ⇒ c = 4
⇒ y =
∴ y(-4) = 0
for critical point
given
∵
= 0 ⇒ ex(x + 3) - 1 = 0⇒ x + 3 = e-x
y(x) has a critical point in the interval (-1, 0)
Create a free account to view solution
View Solution FreeMore Differential Equation Questions
The equation of the curve through the point (1, 0) and whose slope is is-...If φ(x) = φ′(x) and φ(1) = 2, then φ(3) equals...Let $y = y(x)$ be the solution of the differential equation $x\frac{dy}{dx} - y = x^{2}cotx,x \in (0,\pi)$.If $y\left( \...The solution of the differential equation (1 + x2) = x is-...The order of the differential equation of all circles of radius r, having centre on y-axis and passing through the origi...