What is the solution for y^'+tan(x)y=cos^2(x) ?

The solution for y^'+tan(x)y=cos^2(x) is y=(sin(x))/(sec(x))+(c_{1})/(sec(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+tan(x)y=cos^2(x)

y^{^{'}} + tan (x) y = cos^{2} (x)

y = \frac{sin ( x )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

Solution steps

y^{'} (x) + tan (x) y = cos^{2} (x)

Solve linear ODE:

y = \frac{sin ( x )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

y = \frac{sin ( x )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

\frac{\partial}{\partial x} (tan (2 x - y))

\frac{\partial}{\partial x} ((x^{2} - 2) e^{x^{2} y})

\int \frac{30}{30 + e ^{x}} d x

y = 4 (6 - x^{2})^{5}

\int 4 m (9 m^{2} - 10 m) d m

What is the solution for y^'+tan(x)y=cos^2(x) ?
The solution for y^'+tan(x)y=cos^2(x) is y=(sin(x))/(sec(x))+(c_{1})/(sec(x))