y^{''}-6y^'+5y=3x+e^x,y(0)=0,y^'(0)=2

Lösung

y^{^{''}} - 6 y^{^{'}} + 5 y = 3 x + e^{x}, y (0) = 0, y^{^{'}} (0) = 2

y = (\frac{21}{16} - \frac{18}{25}) e^{5 x} - \frac{21}{16} e^{x} + \frac{3 x}{5} + \frac{18}{25} - \frac{e ^{x} x}{4}

Schritte zur Lösung

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

Solve linear ODE:

y = (\frac{21}{16} - \frac{18}{25}) e^{5 x} - \frac{21}{16} e^{x} + \frac{3 x}{5} + \frac{18}{25} - \frac{e ^{x} x}{4}

y = (\frac{21}{16} - \frac{18}{25}) e^{5 x} - \frac{21}{16} e^{x} + \frac{3 x}{5} + \frac{18}{25} - \frac{e ^{x} x}{4}

y^{^{''}} + 2 y^{^{'}} + y = 9 - e^{2 x}, y (0) = 0, y^{^{'}} (0) = 1

(3 d^{2} + 3) y = 3 x

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 + 5 x

y^{^{''}} + 3 y^{^{'}} - 10 y = 20 + 5 e^{3 x}

x^{^{''}} + 4 x^{^{'}} + 14 x = 9 cos (3 t)