y^{''}-y^'-6y=2cos(x)-2e^{3x}+10

Lösung

y^{^{''}} - y^{^{'}} - 6 y = 2 cos (x) - 2 e^{3 x} + 10

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{25} sin (x) - \frac{7}{25} cos (x) - \frac{2 e ^{3 x} x}{5} - \frac{5}{3}

Schritte zur Lösung

y^{''} (x) - y^{'} (x) - 6 y = 2 cos (x) - 2 e^{3 x} + 10

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{25} sin (x) - \frac{7}{25} cos (x) - \frac{2 e ^{3 x} x}{5} - \frac{5}{3}

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{25} sin (x) - \frac{7}{25} cos (x) - \frac{2 e ^{3 x} x}{5} - \frac{5}{3}

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

\frac{d ^{2} x}{d t ^{2}} + 7 \frac{d x}{d t} + 6 x = 5 e^{- t}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 3 cos (t)

y^{^{''}} - 18 y = tan (x)

y^{^{''}} + y = 1, 6 cos (t) y (0) = 0, y^{^{'}} (0) = 0