What is the solution for 2y^{''}+4y=6t ?

The solution for 2y^{''}+4y=6t is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)+(3t)/2

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2y^{''}+4y=6t

2 y^{^{''}} + 4 y = 6 t

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

Solution steps

2 y^{''} (t) + 4 y = 6 t

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 8 y = 8 t + 22, y (0) = - 7, y^{^{'}} (0) = 20

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 x \cdot e^{2 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 e^{t}

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

y^{^{''}} + y = 8 e^{- 6 x}

What is the solution for 2y^{''}+4y=6t ?
The solution for 2y^{''}+4y=6t is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)+(3t)/2