What is the solution for y^{''}+4y=g(t) ?

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

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y^{''}+4y=g(t)

y^{^{''}} + 4 y = g (t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{g t}{4}

Solution steps

y^{''} (t) + 4 y = g (t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{g t}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{g t}{4}

y^{^{''}} - 3 y^{^{'}} = 2 x - 3

y^{^{'''}} - 6 y^{^{''}} + 8 y^{^{'}} = sin (2 t) + t^{2} e^{t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 4

y^{^{''}} + y = 2 x \cdot e^{x} - 2 sin (x)

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

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