What is the solution for y^{''}+25y=3sec^3(5) ?

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

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y^{''}+25y=3sec^3(5)

y^{^{''}} + 25 y = 3 sec^{3} (5)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{3 sec ^{3} ( 5 )}{25}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{3 sec ^{3} ( 5 )}{25}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{3 sec ^{3} ( 5 )}{25}

\frac{x}{x ^{2} + y ^{2}}

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

\frac{d}{d x} (2 x^{- 3})

f (x) = \frac{4}{x}

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

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