What is the solution for cos(t)y^'+sin(t)y-1=0 ?

The solution for cos(t)y^'+sin(t)y-1=0 is y=(tan(t))/(sec(t))+(c_{1})/(sec(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

cos(t)y^'+sin(t)y-1=0

cos (t) y^{^{'}} + sin (t) y - 1 = 0

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

Solution steps

cos (t) y^{'} (t) + sin (t) y - 1 = 0

Solve linear ODE:

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

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

t an g e n t f (x) = 2 x, (1, 2)

\int \frac{4}{x ( x - 4 ) ^{2}} d x

x \to 1 + lim (arcsin (x))

\frac{d}{d x} (e^{2 x} (3 x - 2))

e x p an d (x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1)

What is the solution for cos(t)y^'+sin(t)y-1=0 ?
The solution for cos(t)y^'+sin(t)y-1=0 is y=(tan(t))/(sec(t))+(c_{1})/(sec(t))