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

The solution for y^{''}+y=-sin(2x),y(0)=0,y^'(0)=0 is y=-2/3 sin(x)+1/3 sin(2x)

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y^{''}+y=-sin(2x),y(0)=0,y^'(0)=0

y^{^{''}} + y = - sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{2}{3} sin (x) + \frac{1}{3} sin (2 x)

Solution steps

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

Solve linear ODE:

y = - \frac{2}{3} sin (x) + \frac{1}{3} sin (2 x)

y = - \frac{2}{3} sin (x) + \frac{1}{3} sin (2 x)

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What is the solution for y^{''}+y=-sin(2x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=-sin(2x),y(0)=0,y^'(0)=0 is y=-2/3 sin(x)+1/3 sin(2x)