The solution of the differential equation x s | vedclass

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

Question

(B) Given the differential equation: $x \sec y \frac{dy}{dx} = 1$ Separate the variables: $\sec y \, dy = \frac{1}{x} \, dx$ Integrate both sides: $\i

Vedclass · Accepted Answer

$cx = \sec y + 	an y$

Vedclass · Answer

(B) Given the differential equation: $x \sec y \frac{dy}{dx} = 1$ Separate the variables: $\sec y \, dy = \frac{1}{x} \, dx$ Integrate both sides: $\int \sec y \, dy = \int \frac{1}{x} \, dx$ We know that $\int \sec y \, dy = \log |\sec y + 	an y|$ and $\int \frac{1}{x} \, dx = \log |x| + \log |c|$ So,$\log |\sec y + 	an y| = \log |cx|$ Taking the exponential on both sides,we get: $\sec y + 	an y = cx$

The solution of the differential equation $x \sec y \frac{dy}{dx} = 1$ is

Similar Questions

The solution of $(x - y^3)dx + 3xy^2dy = 0$ is

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

Find the general solution of the differential equation: $x^{5} \frac{dy}{dx} = -y^{5}$

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ is

The solution of the differential equation $x \cos y \, dy = (x e^x \log x + e^x) \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module