What is the solution for y^'=-ytan(t)+(tan(t))/(cos(t)) ?

The solution for y^'=-ytan(t)+(tan(t))/(cos(t)) is y=(tan^2(t))/(2sec(t))+(c_{1})/(sec(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-ytan(t)+(tan(t))/(cos(t))

y^{^{'}} = - y tan (t) + \frac{tan ( t )}{cos ( t )}

y = \frac{tan ^{2} ( t )}{2 sec ( t )} + \frac{c _{1}}{sec ( t )}

Solution steps

y^{'} (t) = - y tan (t) + \frac{tan ( t )}{cos ( t )}

Solve linear ODE:

y = \frac{tan ^{2} ( t )}{2 sec ( t )} + \frac{c _{1}}{sec ( t )}

y = \frac{tan ^{2} ( t )}{2 sec ( t )} + \frac{c _{1}}{sec ( t )}

1 + 3 x, 3 \leq x \leq 9

\frac{d}{d x} (- 27 cos (3 x))

\int \frac{1}{x ^{2} - 2 x + 1} d x

\int cos (\frac{n x π}{2}) d x

\frac{d}{d x} (2 \cdot c \cdot x - x^{2})

What is the solution for y^'=-ytan(t)+(tan(t))/(cos(t)) ?
The solution for y^'=-ytan(t)+(tan(t))/(cos(t)) is y=(tan^2(t))/(2sec(t))+(c_{1})/(sec(t))