What is the solution for y^'=ytan(t)+sin(t) ?

The solution for y^'=ytan(t)+sin(t) is y=(sin^2(t))/(2cos(t))+(c_{1})/(cos(t))

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y^'=ytan(t)+sin(t)

y^{^{'}} = y tan (t) + sin (t)

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

Solution steps

y^{'} (t) = y tan (t) + sin (t)

Solve linear ODE:

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

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

\frac{d}{d x} ((\frac{1}{x} - \frac{4}{x ^{3}}) (x^{3} - 5 x - 1))

x \to 1 lim (x^{5} - 2 x - 6)

\frac{d}{d x} (\frac{csc ( x )}{1 + csc ( x )})

y^{^{''}} + 4 y = e^{4 x}

\int e^{e^{x}} d x

What is the solution for y^'=ytan(t)+sin(t) ?
The solution for y^'=ytan(t)+sin(t) is y=(sin^2(t))/(2cos(t))+(c_{1})/(cos(t))