y^{''}+4y=8t^2

Lösung

y^{^{''}} + 4 y = 8 t^{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 1

Schritte zur Lösung

y^{''} (t) + 4 y = 8 t^{2}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 1

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 2 t^{2} - 1

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

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

5 y^{^{''}} + 6 y^{^{'}} + 80 y = - 24 sin (4 t)

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

y^{^{''}} - 3 y^{^{'}} - 4 y = x