What is the solution for y^{''}+16y=e^{2t}tan(4t) ?

The solution for y^{''}+16y=e^{2t}tan(4t) is

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y^{''}+16y=e^{2t}tan(4t)

y^{^{''}} + 16 y = e^{2 t} tan (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + cos (4 t) \cdot \int - \frac{sin ( 4 t ) e ^{2 t} tan ( 4 t )}{4} d t + \frac{1}{4} (- \frac{e ^{2 t} cos ( 4 t )}{5} + \frac{e ^{2 t} sin ( 4 t )}{10}) sin (4 t)

Solution steps

y^{''} (t) + 16 y = e^{2 t} tan (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + cos (4 t) \cdot \int - \frac{sin ( 4 t ) e ^{2 t} tan ( 4 t )}{4} d t + \frac{1}{4} (- \frac{e ^{2 t} cos ( 4 t )}{5} + \frac{e ^{2 t} sin ( 4 t )}{10}) sin (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + cos (4 t) \cdot \int - \frac{sin ( 4 t ) e ^{2 t} tan ( 4 t )}{4} d t + \frac{1}{4} (- \frac{e ^{2 t} cos ( 4 t )}{5} + \frac{e ^{2 t} sin ( 4 t )}{10}) sin (4 t)

x \to 2 lim ((x^{2} + 8) (2 x - 4))

y \to - \infty lim (\frac{2 - y}{7 + 6 y ^{2}})

\frac{\partial}{\partial y} (26 - x^{2} - 4 y^{2})

\int \frac{6 x}{7 - 3 x ^{2}} d x

\int_{0}^{2} (4 - x^{2})^{2} d x

What is the solution for y^{''}+16y=e^{2t}tan(4t) ?
The solution for y^{''}+16y=e^{2t}tan(4t) is