What is the solution for y^{''}+4y=3tan(2x) ?

The solution for y^{''}+4y=3tan(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(3cos(2x)ln(tan(2x)+sec(2x)))/4

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y^{''}+4y=3tan(2x)

y^{^{''}} + 4 y = 3 tan (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{3 cos ( 2 x ) ln ( tan ( 2 x ) + sec ( 2 x ) )}{4}

Solution steps

y^{''} (x) + 4 y = 3 tan (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{3 cos ( 2 x ) ln ( tan ( 2 x ) + sec ( 2 x ) )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{3 cos ( 2 x ) ln ( tan ( 2 x ) + sec ( 2 x ) )}{4}

\int sin (2 x - 5) d x

\int_{0}^{2 π} 7 x^{2} sin (\frac{1}{4} x) d x

\frac{d y}{d x} = \frac{( y - x )}{x}

n = 2 \sum \infty - \frac{1}{4 ^{n}}

x \to 0 lim (\frac{9 ^{x} - 5 ^{x}}{x})

What is the solution for y^{''}+4y=3tan(2x) ?
The solution for y^{''}+4y=3tan(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-(3cos(2x)ln(tan(2x)+sec(2x)))/4