y^{''}+4y^'=8x+1

解答

y^{^{''}} + 4 y^{^{'}} = 8 x + 1

y = c_{1} + c_{2} e^{- 4 x} + x^{2} - \frac{x}{4}

求解步骤

y^{''} (x) + 4 y^{'} (x) = 8 x + 1

求解线性方程:

y = c_{1} + c_{2} e^{- 4 x} + x^{2} - \frac{x}{4}

y = c_{1} + c_{2} e^{- 4 x} + x^{2} - \frac{x}{4}

y^{^{''}} - y = 2 (e^{x} e^{- x})^{- 1}

y^{^{''}} + 2 y^{^{'}} + y = 3 cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x) - sin (2 x)

x^{^{''}} + 4 x^{^{'}} + 5 x = e^{- 2} t sin (t)

so l v e f or d, d (t) = - 16 t^{2} + 32 t + 40