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

The solution for y^{''}+y=sin(3x),y(0)=0,y^'(0)=0 is y= 3/8 sin(x)-1/8 sin(3x)

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y^{''}+y=sin(3x),y(0)=0,y^'(0)=0

y^{^{''}} + y = sin (3 x), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{3}{8} sin (x) - \frac{1}{8} sin (3 x)

Solution steps

y^{''} (x) + y = sin (3 x)

Solve linear ODE:

y = \frac{3}{8} sin (x) - \frac{1}{8} sin (3 x)

y = \frac{3}{8} sin (x) - \frac{1}{8} sin (3 x)

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

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = - 4

θ^{^{''}} (t) - θ (t) = t sin (t)

y^{^{''}} + 3 y^{^{'}} - 4 y = cos (3 x)

y^{^{''}} + \frac{5}{6} y^{^{'}} + \frac{2}{15} y = \frac{4}{3}

What is the solution for y^{''}+y=sin(3x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=sin(3x),y(0)=0,y^'(0)=0 is y= 3/8 sin(x)-1/8 sin(3x)