What is the solution for y^{''}+y=sin(t),y(0)=0,y^'(0)=2 ?

The solution for y^{''}+y=sin(t),y(0)=0,y^'(0)=2 is y=-(tcos(t))/2+5/2 sin(t)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 6 y = 10 e^{2 x} - 18 e^{3 x} - 6 x - 11

y^{^{''}} + y^{^{'}} = sec^{2} (x)

y^{^{''}} + y^{^{'}} + y = 1 + x

y^{^{'''}} - y^{^{'}} = 3 sin (t)

y^{^{''}} - y = tan^{2} (x)

What is the solution for y^{''}+y=sin(t),y(0)=0,y^'(0)=2 ?
The solution for y^{''}+y=sin(t),y(0)=0,y^'(0)=2 is y=-(tcos(t))/2+5/2 sin(t)