y^{''}-4y^'+8y=x^3+x

Lösung

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3} + x

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{7 x}{32} + \frac{1}{16}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 8 y = x^{3} + x

Solve linear ODE:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{7 x}{32} + \frac{1}{16}

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{x ^{3}}{8} + \frac{3 x ^{2}}{16} + \frac{7 x}{32} + \frac{1}{16}

y^{^{''}} + 4 y = 6 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 3

4 y^{^{''}} - y = 1, y (0) = 0, y^{^{'}} (0) = 0.5

y^{^{''}} + 4 y = 2 x

y^{^{'''}} - 6 y^{^{''}} + 13 y^{^{'}} - 10 y = e^{2 x}

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