y^{''}+3y^'-4y=(x^3+x)e^x

Lösung

y^{^{''}} + 3 y^{^{'}} - 4 y = (x^{3} + x) e^{x}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x ^{4}}{20} - \frac{e ^{x} x ^{3}}{25} + \frac{31 e ^{x} x ^{2}}{250} - \frac{31 e ^{x} x}{625}

Schritte zur Lösung

y^{''} (x) + 3 y^{'} (x) - 4 y = (x^{3} + x) e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x ^{4}}{20} - \frac{e ^{x} x ^{3}}{25} + \frac{31 e ^{x} x ^{2}}{250} - \frac{31 e ^{x} x}{625}

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{e ^{x} x ^{4}}{20} - \frac{e ^{x} x ^{3}}{25} + \frac{31 e ^{x} x ^{2}}{250} - \frac{31 e ^{x} x}{625}

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 e^{3 x} + 36 x^{2}

x^{^{''}} + 16 x = 8 sin (2 t)

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

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

y^{^{''}} + 2 y^{^{'}} + y = 1, y (0) = 0, y^{^{'}} (0) = 0