41\H(x)-8u^'+u^{''}=0

Lösung

41 H (x) - 8 u^{^{'}} + u^{^{''}} = 0

u = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int - \frac{41 H ( x )}{8 e ^{8 x}} d x + \int \frac{41 H ( x )}{8} d x

Schritte zur Lösung

41 H (x) - 8 u^{'} (x) + u^{''} (x) = 0

Solve linear ODE:

u = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int - \frac{41 H ( x )}{8 e ^{8 x}} d x + \int \frac{41 H ( x )}{8} d x

u = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int - \frac{41 H ( x )}{8 e ^{8 x}} d x + \int \frac{41 H ( x )}{8} d x

y^{^{''}} - 3 y = 6 x

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

\frac{1}{2} y^{^{''}} + 5 y^{^{'}} + 12 y = e^{- 4 x} (4 x - 2)

2 y^{^{''}} + 5 y^{^{'}} + 2 y = x^{3} + 9 e^{4 x}

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