What is the solution for y^{''}+36y=4sin(2x),y(0)=1,y^'(0)=0 ?

The solution for y^{''}+36y=4sin(2x),y(0)=1,y^'(0)=0 is y=cos(6x)-1/24 sin(6x)+1/8 sin(2x)

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y^{''}+36y=4sin(2x),y(0)=1,y^'(0)=0

y^{^{''}} + 36 y = 4 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

y = cos (6 x) - \frac{1}{24} sin (6 x) + \frac{1}{8} sin (2 x)

Solution steps

y^{''} (x) + 36 y = 4 sin (2 x)

Solve linear ODE:

y = cos (6 x) - \frac{1}{24} sin (6 x) + \frac{1}{8} sin (2 x)

y = cos (6 x) - \frac{1}{24} sin (6 x) + \frac{1}{8} sin (2 x)

y^{^{''}} + y^{^{'}} - 2 y = 3 - 6 t

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 sin (4 t)

y^{^{''}} - 2 y - 3 y^{^{'}} = 3 t^{2} - 5

y^{^{''}} + y^{^{'}} = cos (3 t)

y^{^{''}} + 2 y^{^{'}} - 35 y = x^{2}

What is the solution for y^{''}+36y=4sin(2x),y(0)=1,y^'(0)=0 ?
The solution for y^{''}+36y=4sin(2x),y(0)=1,y^'(0)=0 is y=cos(6x)-1/24 sin(6x)+1/8 sin(2x)