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

The solution for y^{''}+25y=sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)-(tcos(5t))/(10)

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

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

y = c_{1} cos (5 t) + c_{2} sin (5 t) - \frac{t cos ( 5 t )}{10}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) - \frac{t cos ( 5 t )}{10}

y = c_{1} cos (5 t) + c_{2} sin (5 t) - \frac{t cos ( 5 t )}{10}

x \to 16 lim (x^{2} - 9)

\int \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

\int tan^{5} (1) d x

\frac{\partial}{\partial y} (ln (3 x + 2 y))

x \to 0 + lim ((x + 1)^{l n (x)})

What is the solution for y^{''}+25y=sin(5t) ?
The solution for y^{''}+25y=sin(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)-(tcos(5t))/(10)