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

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

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

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

y = \frac{2 x}{sin ( x )} + \frac{c _{1}}{sin ( x )}

Solution steps

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

Solve linear ODE:

y = \frac{2 x}{sin ( x )} + \frac{c _{1}}{sin ( x )}

y = \frac{2 x}{sin ( x )} + \frac{c _{1}}{sin ( x )}

f (x) = \frac{( x ^{2} )}{( 5 + x )}

f (x) = 4 x^{2} - 2 x + 6, at x = 4

\int_{0}^{1} 7 x 5^{x} d x

\frac{\partial}{\partial y} (x^{2} + x^{2} + y)

\int_{- 1}^{3} (x^{2} - 2 x - 3) d x

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