y^{''}+4y^'=x

Lösung

y^{^{''}} + 4 y^{^{'}} = x

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) = x

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{2}}{8} - \frac{x}{16}

y^{^{''}} + 2 y^{^{'}} - 3 y = (x + 1) e^{x}

y^{^{''}} + y^{^{'}} = 4 cos (2 x)

6 y^{^{''}} - 5 y^{^{'}} + y = 3 x^{2} + x

y^{^{''}} + y^{^{'}} - 6 y = 5 x e^{3 x}

y^{^{''}} - 6 y + 9 y = t^{2} e^{3 t}