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

解

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

解答ステップ

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

線形 ODE を解く:

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