What is the solution for (dy)/(dx)-4tan(2x)=sin(2x) ?

The solution for (dy)/(dx)-4tan(2x)=sin(2x) is y=-1/2 cos(2x)-2ln(cos(2x))+c_{1}

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(dy)/(dx)-4tan(2x)=sin(2x)

\frac{d y}{d x} - 4 tan (2 x) = sin (2 x)

y = - \frac{1}{2} cos (2 x) - 2 ln (cos (2 x)) + c_{1}

Solution steps

\frac{d y}{d x} - 4 tan (2 x) = sin (2 x)

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int sin (2 x) + 4 tan (2 x) d x

\int sin (2 x) + 4 tan (2 x) d x = - \frac{1}{2} cos (2 x) - 2 ln (cos (2 x)) + c_{1}

y = - \frac{1}{2} cos (2 x) - 2 ln (cos (2 x)) + c_{1}

x \to - 3 lim (- 3)

\int_{0}^{1} \frac{1}{( 2 x + 1 ) ^{3}} d x

V (r) = (\frac{4}{3}) π r^{3}

\frac{d y}{d t} = 3 t^{2} y + (4 t + 1) e^{t^{3}}

\frac{\partial}{\partial z} (sin (x z))

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