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

The solution for y^{''}+y=sin(t)cos(t) is y=c_{1}cos(t)+c_{2}sin(t)-(cos(t)sin(t))/3

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

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

y = c_{1} cos (t) + c_{2} sin (t) - \frac{cos ( t ) sin ( t )}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - \frac{cos ( t ) sin ( t )}{3}

y = c_{1} cos (t) + c_{2} sin (t) - \frac{cos ( t ) sin ( t )}{3}

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( 2 e ^{- 3 x} )}{x ^{2} + 1}

y^{^{''}} + y = x^{2} + 1, y (0) = 7, y (1) = 0

8 y^{^{''}} + 2 y^{^{'}} - y = 9

y^{^{''}} - 8 y^{^{'}} + 15 y = 64 e^{t}

d (x) = (3 x - 1) (x - 2)^{- 1}

What is the solution for y^{''}+y=sin(t)cos(t) ?
The solution for y^{''}+y=sin(t)cos(t) is y=c_{1}cos(t)+c_{2}sin(t)-(cos(t)sin(t))/3