What is the solution for y^{''}+9y=cos(2t) ?

The solution for y^{''}+9y=cos(2t) is y=c_{1}cos(3t)+c_{2}sin(3t)+1/5 cos(2t)

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y^{''}+9y=cos(2t)

y^{^{''}} + 9 y = cos (2 t)

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

Solution steps

y^{''} (t) + 9 y = cos (2 t)

Solve linear ODE:

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

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

h \to 0 lim (\frac{( ( 4 + h ) - 4 )}{h})

\frac{d}{d x} (a x e^{4 x})

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

\int 12 x z cos (x z^{2}) d z

f (x) = ln (1 + sin^{2} (x))

What is the solution for y^{''}+9y=cos(2t) ?
The solution for y^{''}+9y=cos(2t) is y=c_{1}cos(3t)+c_{2}sin(3t)+1/5 cos(2t)