Question:

If $(2 + \sin \, x ) \frac{dy}{dx} + (y + 1) \cos \, x = 0$ and $y(0) = 1,$ then $y \left( \frac{\pi}{2} \right)$ is equal to :

Updated On: Sep 30, 2024
  • $- \frac{2}{3}$
  • $- \frac{1}{3}$
  • $ \frac{4}{3}$
  • $ \frac{1}{3}$
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The Correct Option is D

Solution and Explanation

$\left(2+\sin x\right) \frac{dy}{dx}+\left(y+1\right) \cos x = 0$
$y (0) = 1, y \left(\frac{\pi}{2}\right)$ = ?
$\frac{1}{y+1}dy+\frac{\cos x}{2+\sin x} dx = 0$
$In\left|y+1\right|+In\left(2+\sin x\right) = InC$
$(y + 1) (2 + \sin x) = C$
Put $x = 0,\, y = 1$
$(1 + 1) \cdot 2 = C \Rightarrow C = 4$
Now, $(y +1)(2 + \sin x) = 4$
For, $x = \frac{\pi}{2}$
$(y +1)(2 +1) = 4$
$y + 1 = \frac{4}{3}$
$y = \frac{4}{3}-1 =\frac{1}{3}$
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Concepts Used:

Differential Equations

A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of the function define the rate of change of a function at a point. It is mainly used in fields such as physics, engineering, biology and so on.

Orders of a Differential Equation

First Order Differential Equation

The first-order differential equation has a degree equal to 1. All the linear equations in the form of derivatives are in the first order. It has only the first derivative such as dy/dx, where x and y are the two variables and is represented as: dy/dx = f(x, y) = y’

Second-Order Differential Equation

The equation which includes second-order derivative is the second-order differential equation. It is represented as; d/dx(dy/dx) = d2y/dx2 = f”(x) = y”.

Types of Differential Equations

Differential equations can be divided into several types namely

  • Ordinary Differential Equations
  • Partial Differential Equations
  • Linear Differential Equations
  • Nonlinear differential equations
  • Homogeneous Differential Equations
  • Nonhomogeneous Differential Equations