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

Lösung

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

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

8 y^{^{''}} - 32 y = 16 cos (x)

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

\frac{d ^{2} y}{d x ^{2}} + y = sin (2 x)

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} - y^{^{'}} - 6 y = (x^{2} + 3 x) e^{2 x} + cos (2 x)