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)

x^{^{''}} + 4 x = 289 + e^{t} \cdot sin (2 t)

y^{^{''}} - 10 y^{^{'}} + 125 y = - 4 sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + y = 2 x sin (x)

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

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 10 x^{2} - 8 x - 15

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)