y^{''}+2y^'-3y=tan(x)

Lösung

y^{^{''}} + 2 y^{^{'}} - 3 y = tan (x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + e^{x} \cdot \int \frac{tan ( x )}{4 e ^{x}} d x + e^{- 3 x} \cdot \int - \frac{e ^{3 x} tan ( x )}{4} d x

Schritte zur Lösung

y^{''} (x) + 2 y^{'} (x) - 3 y = tan (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + e^{x} \cdot \int \frac{tan ( x )}{4 e ^{x}} d x + e^{- 3 x} \cdot \int - \frac{e ^{3 x} tan ( x )}{4} d x

y = c_{1} e^{x} + c_{2} e^{- 3 x} + e^{x} \cdot \int \frac{tan ( x )}{4 e ^{x}} d x + e^{- 3 x} \cdot \int - \frac{e ^{3 x} tan ( x )}{4} d x

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 5 y = e^{x} cos (2 x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 2 e^{- 2 x} + 3

x^{^{''}} - 4 x^{^{'}} + 4 x = e^{2 t}, x (0) = 2, x^{^{'}} (0) = 1

y^{^{'''}} + 256 y^{^{'}} = tan (16 x)

y^{^{''}} - 4 y = 3 x^{2} e^{3 x}