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

The solution for y^{''}+y=sin(x),y(0)=0,y^'(0)=2 is y=-(xcos(x))/2+5/2 sin(x)

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

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

y = - \frac{x cos ( x )}{2} + \frac{5}{2} sin (x)

Solution steps

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

Solve linear ODE:

y = - \frac{x cos ( x )}{2} + \frac{5}{2} sin (x)

y = - \frac{x cos ( x )}{2} + \frac{5}{2} sin (x)

y^{^{'''}} + 169 y^{^{'}} = tan (13 x)

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

y^{^{''}} + 16 y = 6 cos (4 x) + 9 sin (4 x)

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 = 0

y^{^{''''}} - y = 12 t + cos (t)

What is the solution for y^{''}+y=sin(x),y(0)=0,y^'(0)=2 ?
The solution for y^{''}+y=sin(x),y(0)=0,y^'(0)=2 is y=-(xcos(x))/2+5/2 sin(x)