y^{''}+y^'+2y=-25xe^x,y^'(0)=0

解

y^{^{''}} + y^{^{'}} + 2 y = - 25 x e^{x}, y^{^{'}} (0) = 0

y = e^{- \frac{x}{2}} - \frac{cos ( \frac{7 x}{2} ) ( - 8 7 c _{2} + 25 )}{8} + c_{2} sin (\frac{7 x}{2}) - \frac{25 e ^{x} x}{4} + \frac{75}{16} e^{x}

解答ステップ

y^{''} (x) + y^{'} (x) + 2 y = - 25 x e^{x}

線形 ODE を解く:

y = e^{- \frac{x}{2}} - \frac{cos ( \frac{7 x}{2} ) ( - 8 7 c _{2} + 25 )}{8} + c_{2} sin (\frac{7 x}{2}) - \frac{25 e ^{x} x}{4} + \frac{75}{16} e^{x}

y = e^{- \frac{x}{2}} - \frac{cos ( \frac{7 x}{2} ) ( - 8 7 c _{2} + 25 )}{8} + c_{2} sin (\frac{7 x}{2}) - \frac{25 e ^{x} x}{4} + \frac{75}{16} e^{x}