y^{''}+2y^'-15y=(2x+1)e^{3x}+10

Lösung

y^{^{''}} + 2 y^{^{'}} - 15 y = (2 x + 1) e^{3 x} + 10

y = c_{1} e^{3 x} + c_{2} e^{- 5 x} - \frac{2}{3} + \frac{4 e ^{3 x} x ^{2} + 3 e ^{3 x} x}{32}

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) - 15 y = (2 x + 1) e^{3 x} + 10

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 5 x} - \frac{2}{3} + \frac{4 e ^{3 x} x ^{2} + 3 e ^{3 x} x}{32}

y = c_{1} e^{3 x} + c_{2} e^{- 5 x} - \frac{2}{3} + \frac{4 e ^{3 x} x ^{2} + 3 e ^{3 x} x}{32}

y^{^{''}} - y^{^{'}} - 2 y = 2 t^{2} - 5

y^{^{''}} - 3 y^{^{'}} - 4 y = 2^{t - 2}, y (2) = 1, y^{^{'}} (2) = - 1

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

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

y^{^{''}} - y = \frac{2 e ^{2 x}}{e ^{2 x} + 1}