What is the solution for y^'+tan(x)y-sin(x)=0 ?

The solution for y^'+tan(x)y-sin(x)=0 is y=-(ln(cos(x)))/(sec(x))+(c_{1})/(sec(x))

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y^'+tan(x)y-sin(x)=0

y^{^{'}} + tan (x) y - sin (x) = 0

y = - \frac{ln ( cos ( x ) )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

Solution steps

y^{'} (x) + tan (x) y - sin (x) = 0

Solve linear ODE:

y = - \frac{ln ( cos ( x ) )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

y = - \frac{ln ( cos ( x ) )}{sec ( x )} + \frac{c _{1}}{sec ( x )}

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

\frac{\partial}{\partial x} (4 sin (4 x))

\frac{d}{d x} (arccos (x + y))

f (x) = 3 x^{2}, at x = (6)

x \to 1 + lim (\frac{1}{x ^{3} - 1})

What is the solution for y^'+tan(x)y-sin(x)=0 ?
The solution for y^'+tan(x)y-sin(x)=0 is y=-(ln(cos(x)))/(sec(x))+(c_{1})/(sec(x))