y^{''}+4y^'+20y=-3sin(2x)

Lösung

y^{^{''}} + 4 y^{^{'}} + 20 y = - 3 sin (2 x)

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) - \frac{3}{20} sin (2 x) + \frac{3}{40} cos (2 x)

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 20 y = - 3 sin (2 x)

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) - \frac{3}{20} sin (2 x) + \frac{3}{40} cos (2 x)

y = e^{- 2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) - \frac{3}{20} sin (2 x) + \frac{3}{40} cos (2 x)

\frac{d}{d x} (\frac{ln ( x ^{2} + 1 )}{3 x + 1})

1 + ln (x) + \frac{y}{x} d x = (8 - ln (x)) d y

\frac{d}{d x} (x sin (x + y))

\int \frac{x}{x ^{2} + 2 x + 25} d x

\int \frac{5 x + 7}{4 x ^{2} + 12 x + 10} d x