What is the solution for y^'-4tan(2x)y=sin(2x) ?

The solution for y^'-4tan(2x)y=sin(2x) is y=-(cos(2x))/6+(c_{1})/(cos^2(2x))

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y^'-4tan(2x)y=sin(2x)

y^{^{'}} - 4 tan (2 x) y = sin (2 x)

y = - \frac{cos ( 2 x )}{6} + \frac{c _{1}}{cos ^{2} ( 2 x )}

Solution steps

y^{'} (x) - 4 tan (2 x) y = sin (2 x)

Solve linear ODE:

y = - \frac{cos ( 2 x )}{6} + \frac{c _{1}}{cos ^{2} ( 2 x )}

y = - \frac{cos ( 2 x )}{6} + \frac{c _{1}}{cos ^{2} ( 2 x )}

\int_{0}^{1} \frac{24}{x ^{2} - 36} d x

\int 10 e^{3 x} d x

\frac{d}{d x} (x + 8)

x \to 1 - lim (\frac{2 x + 5}{x + 1})

x \to 0 + lim (ln (\frac{1}{2 x}))

What is the solution for y^'-4tan(2x)y=sin(2x) ?
The solution for y^'-4tan(2x)y=sin(2x) is y=-(cos(2x))/6+(c_{1})/(cos^2(2x))