y^{''}-6y^'+8y=xe^{3x}cos(4x)

Lösung

y^{^{''}} - 6 y^{^{'}} + 8 y = x e^{3 x} cos (4 x)

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

Schritte zur Lösung

y^{''} (x) - 6 y^{'} (x) + 8 y = x e^{3 x} cos (4 x)

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} - 20 y = 5 e^{- 2 x}

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

y^{^{''}} + 9 y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 6, y^{^{'}} (0) = 0

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