The solution of the differential equation cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the differential equation: $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ Separate the variables $x$ and $y$: $\frac{\cos y}{\sin y} dy = -

Vedclass · Accepted Answer

$\sin y = c \cos x$

Vedclass · Answer

(D) Given the differential equation: $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ Separate the variables $x$ and $y$: $\frac{\cos y}{\sin y} dy = - \frac{\sin x}{\cos x} dx$ This simplifies to: $\cot y \, dy = - 	an x \, dx$ Integrating both sides: $\int \cot y \, dy = - \int 	an x \, dx$ Using the standard integrals $\int \cot y \, dy = \log |\sin y|$ and $\int 	an x \, dx = \log |\sec x| = - \log |\cos x|$,we get: $\log |\sin y| = - (- \log |\cos x|) + \log |c|$ $\log |\sin y| = \log |\cos x| + \log |c|$ Using the property $\log a + \log b = \log(ab)$: $\log |\sin y| = \log |c \cos x|$ Taking the exponential of both sides: $\sin y = c \cos x$

The solution of the differential equation $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ is

Similar Questions

The solution of ${e^{2x - 3y}}dx + {e^{2y - 3x}}dy = 0$ is

The solution of the differential equation $\left( {1 + {e^{2y}}} \right){e^{{{\tan }^{ - 1}}x}}dx - \left( {1 + {x^2}} \right)\left( {{e^y} + {{\left( {{e^y} - 1} \right)}^2}} \right)dy = 0$ is

The solution of the differential equation $(x + y - 1)dx + (2x + 2y - 3)dy = 0$ is

If $\frac{dy}{dx} + 2x \tan(x-y) = 1$, then $\sin(x-y)$ is equal to

Solve the differential equation $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module