What is the solution for y^{''}+36y=33cos(5t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+36y=33cos(5t),y(0)=0,y^'(0)=0 is y=3cos(5t)-3cos(6t)

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y^{''}+36y=33cos(5t),y(0)=0,y^'(0)=0

y^{^{''}} + 36 y = 33 cos (5 t), y (0) = 0, y^{^{'}} (0) = 0

y = 3 cos (5 t) - 3 cos (6 t)

Solution steps

y^{''} (t) + 36 y = 33 cos (5 t)

Solve linear ODE:

y = 3 cos (5 t) - 3 cos (6 t)

y = 3 cos (5 t) - 3 cos (6 t)

D (q) = 2300 - 42 q

y^{^{''}} + 2 y^{^{'}} = x

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x - 1 e x

y^{^{''}} - 36 y = - 108 t + 72 t^{2}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 4 e^{x} - 9

What is the solution for y^{''}+36y=33cos(5t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+36y=33cos(5t),y(0)=0,y^'(0)=0 is y=3cos(5t)-3cos(6t)