For the differential equations, find the general solution:\(\frac{dy}{dx}=\frac{1-cosx}{1+cosx}\)
Solution and Explanation
The given differential equation is:
\(\frac{dy}{dx}=\frac{1-cosx}{1+cosx}\)
\(⇒\frac{dy}{dx}={2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}=tan^2\frac{x}{2}\)
\(⇒\frac{dy}{dx}=(sec^2\frac{x}{2}-1)\)
separating the variables,we get:
\(dy=(sec^2\frac{x}{2}-1)dx\)
Now,integrating both sides of this equation,we get:
\(∫dy=∫(sec^2\frac{x}{2}-1)dx=∫sec^2\frac{x}{2}dx-∫dx\)
\(⇒y=2tan\frac{x}{2}-x+C\)
This is the required general solution of the given differential equation.
Top Questions on Differential equations
- If \( x = f(y) \) is the solution of the differential equation \[ (1 + y^2) + (x - 2e^{\tan^{-1}y}) \frac{dy}{dx} = 0, \quad y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \] with \( f(0) = 1 \), then \( f\left( \frac{1}{\sqrt{3}} \right) \) is equal to:
- JEE Main - 2025
- Mathematics
- Differential equations
- Let \( x = x(y) \) be the solution of the differential equation \( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \). If \( x(1) = 1 \), then \( x\left( \frac{1}{2} \right) \) is :
- JEE Main - 2025
- Mathematics
- Differential equations
- Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to __________.
- JEE Main - 2025
- Mathematics
- Differential equations
- If a curve $ y = y(x) $ passes through the point $ \left(1, \frac{\pi}{2}\right) $ and satisfies the differential equation $$ (7x^4 \cot y - e^x \csc y) \frac{dx}{dy} = x^5, \quad x \geq 1, \text{ then at } x = 2, \text{ the value of } \cos y \text{ is:} $$
- JEE Main - 2025
- Mathematics
- Differential equations
- Let \( y = y(x) \) be the solution of the differential equation \[ \cos(x \log(\cos x))^2 \, dy + (\sin x - 3 \sin x \log(\cos x)) \, dx = 0, \quad x \in \left( 0, \frac{\pi}{2} \right) \] If \( y\left( \frac{\pi}{4} \right) = -1 \), then \( y\left( \frac{\pi}{6} \right) \) is equal to:
- JEE Main - 2025
- Mathematics
- Differential equations
Questions Asked in CBSE CLASS XII exam
- Find the value of $x$, if \[ \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ x \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]
- Two point charges of \( -5\,\mu C \) and \( 2\,\mu C \) are located in free space at \( (-4\,\text{cm}, 0) \) and \( (6\,\text{cm}, 0) \) respectively.
(a) Calculate the amount of work done to separate the two charges at infinite distance.
(b) If this system of charges was initially kept in an electric field \[ \vec{E} = \frac{A}{r^2}, \text{ where } A = 8 \times 10^4\, \text{N}\,\text{C}^{-1}\,\text{m}^2, \] calculate the electrostatic potential energy of the system.- CBSE CLASS XII - 2025
- Electrostatics
- 4,000 shares of ₹ 10 each were forfeited for non-payment of second and final call money of ₹ 2 per share. The minimum amount that the company must collect at the time of reissue of these shares will be :
- CBSE CLASS XII - 2025
- Accounting for Share Capital
- If $y = a \cos(\log x) + b \sin(\log x)$, then $x^2y'' + xy'1$ is:
- CBSE CLASS XII - 2025
- Continuity and differentiability
A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?
- CBSE CLASS XII - 2025
- Application of derivatives
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