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

Lösung

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{e ^{- 2 x} arccos ( 2 x )}{4} d x + e^{- 2 x} \cdot \int - \frac{e ^{2 x} arccos ( 2 x )}{4} d x

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{e ^{- 2 x} arccos ( 2 x )}{4} d x + e^{- 2 x} \cdot \int - \frac{e ^{2 x} arccos ( 2 x )}{4} d x

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{e ^{- 2 x} arccos ( 2 x )}{4} d x + e^{- 2 x} \cdot \int - \frac{e ^{2 x} arccos ( 2 x )}{4} d x

y^{^{''}} + y^{^{'}} = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

l in e a r y^{^{''}} + 4 y^{^{'}} + 4 y = t - 2 e - 2 t

y^{^{''}} + y^{^{'}} = 2 sin (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 11 e^{- 3 x}

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 e^{- 2 x}