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

The solution for y^{''}+b^2y=sin(bt),y(0)=0,y^'(0)=0 is y=-(tcos(bt))/(2b)+(sin(bt))/(2b^2)

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

y^{^{''}} + b^{2} y = sin (b t), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{t cos ( b t )}{2 b} + \frac{sin ( b t )}{2 b ^{2}}

Solution steps

y^{''} (t) + b^{2} y = sin (b t)

Solve linear ODE:

y = - \frac{t cos ( b t )}{2 b} + \frac{sin ( b t )}{2 b ^{2}}

y = - \frac{t cos ( b t )}{2 b} + \frac{sin ( b t )}{2 b ^{2}}

What is the solution for y^{''}+b^2y=sin(bt),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+b^2y=sin(bt),y(0)=0,y^'(0)=0 is y=-(tcos(bt))/(2b)+(sin(bt))/(2b^2)