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

The solution for y^{''}+4y=sin(x),y(0)=1,y^'(0)=2 is y=cos(2x)+5/6 sin(2x)+1/3 sin(x)

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

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

y = cos (2 x) + \frac{5}{6} sin (2 x) + \frac{1}{3} sin (x)

Solution steps

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

Solve linear ODE:

y = cos (2 x) + \frac{5}{6} sin (2 x) + \frac{1}{3} sin (x)

y = cos (2 x) + \frac{5}{6} sin (2 x) + \frac{1}{3} sin (x)

What is the solution for y^{''}+4y=sin(x),y(0)=1,y^'(0)=2 ?
The solution for y^{''}+4y=sin(x),y(0)=1,y^'(0)=2 is y=cos(2x)+5/6 sin(2x)+1/3 sin(x)