What is the solution for y^{''}+25y=sec^3(5t) ?

The solution for y^{''}+25y=sec^3(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+(tan(5t)sin(5t))/(50)

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y^{''}+25y=sec^3(5t)

y^{^{''}} + 25 y = sec^{3} (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{tan ( 5 t ) sin ( 5 t )}{50}

Solution steps

y^{''} (t) + 25 y = sec^{3} (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{tan ( 5 t ) sin ( 5 t )}{50}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{tan ( 5 t ) sin ( 5 t )}{50}

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

y^{^{''}} + 4 y = 1 - t^{3}

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

y^{^{''}} + y = sin^{2} (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = 1 + x - 7 e^{- 2 x}

What is the solution for y^{''}+25y=sec^3(5t) ?
The solution for y^{''}+25y=sec^3(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+(tan(5t)sin(5t))/(50)