What is the solution for y^{''}+25y=-40sin(5t) ?

The solution for y^{''}+25y=-40sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+4tcos(5t)

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y^{''}+25y=-40sin(5t)

y^{^{''}} + 25 y = - 40 sin (5 t)

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

Solution steps

y^{''} (t) + 25 y = - 40 sin (5 t)

Solve linear ODE:

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

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

x^{2} y^{^{'}} - 2 x y = y^{4}

\frac{d}{d x} (cos^{2} (tan (x)))

x \to 0 lim (\frac{5 x}{x ^{3}})

\frac{d}{d x} (\frac{x - 1}{x ^{2} + 2 x + 6})

y^{^{'}} - 2 y = - 5 x^{2}

What is the solution for y^{''}+25y=-40sin(5t) ?
The solution for y^{''}+25y=-40sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)+4tcos(5t)