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

Lösung

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

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{3}}{3} + \frac{1}{2} e^{x} x^{2} + \frac{- e ^{2} x ^{2} - 2 e ^{2}}{2} - \frac{3}{2} e^{x} x - 2 x + \frac{7}{4} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{3}}{3} + \frac{1}{2} e^{x} x^{2} + \frac{- e ^{2} x ^{2} - 2 e ^{2}}{2} - \frac{3}{2} e^{x} x - 2 x + \frac{7}{4} e^{x}

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{3}}{3} + \frac{1}{2} e^{x} x^{2} + \frac{- e ^{2} x ^{2} - 2 e ^{2}}{2} - \frac{3}{2} e^{x} x - 2 x + \frac{7}{4} e^{x}

y^{^{''}} + y - 2 y = 2 t - 40 cos (2 t)

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

y^{^{''}} - 2 y^{^{'}} = cos (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 11 y^{^{'}} + 18 y = 6

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = x^{2} - 2 x