y^{''}+4y=x^2+3e^x,y(0)=0,y^'(0)=2

解答

y^{^{''}} + 4 y = x^{2} + 3 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

y = - \frac{19}{40} cos (2 x) + \frac{7}{10} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{x}

求解步骤

y^{''} (x) + 4 y = x^{2} + 3 e^{x}

求解线性方程:

y = - \frac{19}{40} cos (2 x) + \frac{7}{10} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{x}

y = - \frac{19}{40} cos (2 x) + \frac{7}{10} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{3}{5} e^{x}