What is the solution for y^{''}+25y=cos(4t) ?

The solution for y^{''}+25y=cos(4t) is y=c_{1}cos(5t)+c_{2}sin(5t)+1/9 cos(4t)

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y^{''}+25y=cos(4t)

y^{^{''}} + 25 y = cos (4 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{9} cos (4 t)

Solution steps

y^{''} (t) + 25 y = cos (4 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{9} cos (4 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{1}{9} cos (4 t)

x \to 0 lim (2 x cot (3 x))

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

(x + 2)^{2} + (y - 3)^{2} = 37, (- 1, - 3)

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

y = \frac{x + x ^{3}}{x ^{2}}

What is the solution for y^{''}+25y=cos(4t) ?
The solution for y^{''}+25y=cos(4t) is y=c_{1}cos(5t)+c_{2}sin(5t)+1/9 cos(4t)