What is the solution for cos(x)y^'+ysin(x)=1 ?

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

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cos(x)y^'+ysin(x)=1

cos (x) y^{^{'}} + y sin (x) = 1

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

Solution steps

cos (x) y^{'} (x) + y sin (x) = 1

Solve linear ODE:

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

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

\int cos^{4} (5 x + 8) d x

n = 1 \sum \infty \frac{n}{3 ^{n}}

\frac{\partial}{\partial θ} (ln (θ + 1))

\int \frac{( e ^{t a n (x)} )}{cos ^{2} ( x )} d x

y^{^{''}} + 5 y^{^{'}} + 2 y = 7

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