y^{'''}+y^'=4x

解答

y^{^{'''}} + y^{^{'}} = 4 x

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

求解步骤

y^{'''} (x) + y^{'} (x) = 4 x

求解线性方程:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) + 2 x^{2}

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

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 40 y = - 132 t + 80 t^{2}

y^{^{''}} + 8 y^{^{'}} + 12 = 0

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 e^{t} + 2 t + 2