The general solution of the differential equa | vedclass

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

Question

(B) Given the differential equation: $\frac{dy}{dx} = \cot x \cot y$ Separate the variables: $\frac{dy}{\cot y} = \cot x \, dx$ This can be written as

Vedclass · Accepted Answer

$\sin x = c \sec y$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = \cot x \cot y$ Separate the variables: $\frac{dy}{\cot y} = \cot x \, dx$ This can be written as: $	an y \, dy = \cot x \, dx$ Integrate both sides: $\int 	an y \, dy = \int \cot x \, dx$ Using the standard integration formulas $\int 	an y \, dy = \ln|\sec y|$ and $\int \cot x \, dx = \ln|\sin x|$: $\ln|\sec y| = \ln|\sin x| + C$ Let the constant $C = \ln|c|$: $\ln|\sec y| = \ln|\sin x| + \ln|c|$ Using the property $\ln a + \ln b = \ln(ab)$: $\ln|\sec y| = \ln|c \sin x|$ Taking the exponential of both sides: $\sec y = c \sin x$ This can be rewritten as: $\sin x = \frac{1}{c} \sec y$ Since $\frac{1}{c}$ is also an arbitrary constant,we can write it as $c$: $\sin x = c \sec y$

The general solution of the differential equation $\frac{dy}{dx} = \cot x \cot y$ is

Similar Questions

If the solution $y=y(x)$ of the differential equation $(x^4+2x^3+3x^2+2x+2)dy-(2x^2+2x+3)dx=0$ satisfies $y(-1)=-\frac{\pi}{4}$,then $y(0)$ is equal to:

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$,then $f(xy)$ is equal to

Find a particular solution satisfying the given condition:
$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x; y=1$ when $x=0$

The solution of $(x+y)^{2} \frac{dy}{dx} = a^{2}$ (where $a$ is a constant) is:

If $2x - y + c \log(x - 2y - 4) = k$ is the general solution of $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$,then $c =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module