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

The solution for y^{''}+25y=tsin(4t) is y=c_{1}cos(5t)+c_{2}sin(5t)+(tsin(4t))/9-8/81 cos(4t)

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

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

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 4 t )}{9} - \frac{8}{81} cos (4 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 4 t )}{9} - \frac{8}{81} cos (4 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{t sin ( 4 t )}{9} - \frac{8}{81} cos (4 t)

\frac{\partial}{\partial w} (\frac{w}{u})

x \to 0 lim (\frac{e ^{x} - x - 1}{x})

8^{x}

\frac{d}{d x} (\frac{cot ( x )}{1 - sin ( x )})

\frac{d}{d x} (sin (8 x) e^{4 x})

What is the solution for y^{''}+25y=tsin(4t) ?
The solution for y^{''}+25y=tsin(4t) is y=c_{1}cos(5t)+c_{2}sin(5t)+(tsin(4t))/9-8/81 cos(4t)