y^{'''}+y^'-10=xe^{2x}

解答

y^{^{'''}} + y^{^{'}} - 10 = x e^{2 x}

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{2 x} x}{10} - \frac{13}{100} e^{2 x} + 10 x

求解步骤

y^{'''} (x) + y^{'} (x) - 10 = x e^{2 x}

求解线性方程:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{2 x} x}{10} - \frac{13}{100} e^{2 x} + 10 x

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + \frac{e ^{2 x} x}{10} - \frac{13}{100} e^{2 x} + 10 x