y^{''}-2y^'+y=(e/x)^x

解

y^{^{''}} - 2 y^{^{'}} + y = (\frac{e}{x})^{x}

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

解答ステップ

y^{''} (x) - 2 y^{'} (x) + y = (\frac{e}{x})^{x}

線形 ODE を解く:

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

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