What is the solution for y^{''}+25y=64cos(3x),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+25y=64cos(3x),y(0)=0,y^'(0)=0 is y=4cos(3x)-4cos(5x)

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y^{''}+25y=64cos(3x),y(0)=0,y^'(0)=0

y^{^{''}} + 25 y = 64 cos (3 x), y (0) = 0, y^{^{'}} (0) = 0

y = 4 cos (3 x) - 4 cos (5 x)

Solution steps

y^{''} (x) + 25 y = 64 cos (3 x)

Solve linear ODE:

y = 4 cos (3 x) - 4 cos (5 x)

y = 4 cos (3 x) - 4 cos (5 x)

y^{^{''}} + y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

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

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

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 30, y (0) = 0, y^{^{'}} (0) = 0

q^{^{''}} + q^{^{'}} + \frac{1}{0.05} = 80 t

What is the solution for y^{''}+25y=64cos(3x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+25y=64cos(3x),y(0)=0,y^'(0)=0 is y=4cos(3x)-4cos(5x)