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

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

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y^'=y*tan(x)+sin(x)

y^{^{'}} = y \cdot tan (x) + sin (x)

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

Solution steps

y^{'} (x) = y \cdot tan (x) + sin (x)

Solve linear ODE:

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

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

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

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f (x) = \frac{( x ^{3} + 9 )}{x ^{3}}

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