y^{''}-8y^'+16=10e^{4x}sec(x)

解答

y^{^{''}} - 8 y^{^{'}} + 16 = 10 e^{4 x} sec (x)

y = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int \frac{5 e ^{4 x} sec ( x ) - 8}{4 e ^{8 x}} d x + \int - \frac{5 e ^{4 x} sec ( x ) - 8}{4} d x

求解步骤

y^{''} (x) - 8 y^{'} (x) + 16 = 10 e^{4 x} sec (x)

求解线性方程:

y = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int \frac{5 e ^{4 x} sec ( x ) - 8}{4 e ^{8 x}} d x + \int - \frac{5 e ^{4 x} sec ( x ) - 8}{4} d x

y = c_{1} e^{8 x} + c_{2} + e^{8 x} \cdot \int \frac{5 e ^{4 x} sec ( x ) - 8}{4 e ^{8 x}} d x + \int - \frac{5 e ^{4 x} sec ( x ) - 8}{4} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3 t} ln (t)

\frac{d ^{2} x}{d t ^{2}} + 5 \frac{d x}{d t} + 3 x = t + 1

y^{^{''}} - 2 y^{^{'}} + y = 4 x e^{x}

y^{^{''}} + 3 y^{^{'}} = 10

y^{^{''}} - y = 3 x^{2} + 1