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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{x} + 7

(D^{2} - D - 2) y = 36 x e^{2 x}

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 sin (x) + (12 x - 10) e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} - 4 y = 2 sin (x)

y^{^{''}} + 16 y = - 8 sin (4 x) + 17 e^{x}

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