What is the solution for tan^2(x)cos^2(y)dx=dy ?

The solution for tan^2(x)cos^2(y)dx=dy is y=arctan(-x+tan(x)+c_{1})+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(x)cos^2(y)dx=dy

tan^{2} (x) cos^{2} (y) d x = d y

y = arctan (- x + tan (x) + c_{1}) + πn

Solution steps

tan^{2} (x) cos^{2} (y) d x = d y

Solve separable ODE:

y = arctan (- x + tan (x) + c_{1}) + πn

y = arctan (- x + tan (x) + c_{1}) + πn

\frac{d y}{d x} = \frac{- 6}{4} x^{5} y^{8}

\frac{d y}{d x} = \frac{2 y}{x + 3}

x \frac{d y}{d x} = \frac{1 - 4 y ^{2}}{3 y}

\frac{d y}{d x} y = 5^{x}

\frac{d v}{d t} = v^{2} + 16

What is the solution for tan^2(x)cos^2(y)dx=dy ?
The solution for tan^2(x)cos^2(y)dx=dy is y=arctan(-x+tan(x)+c_{1})+pin