y^{''}-4y^'+3y=3x^2-5x+1

解答

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 x^{2} - 5 x + 1

y = c_{1} e^{3 x} + c_{2} e^{x} + x^{2} + x + 1

求解步骤

y^{''} (x) - 4 y^{'} (x) + 3 y = 3 x^{2} - 5 x + 1

求解线性方程:

y = c_{1} e^{3 x} + c_{2} e^{x} + x^{2} + x + 1

y = c_{1} e^{3 x} + c_{2} e^{x} + x^{2} + x + 1

\int \frac{4 x ^{2} + 2}{1 + x ^{2}} d x

\int_{1}^{5} \frac{3}{x ^{2}} d x

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 3

\frac{\partial}{\partial x} (2 (x - y) e^{2 y + x^{2}})