(dy)/(dt)= y/(1+t)-y/t+t^2(1+t)

解答

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

y = \frac{t ^{3} ( t + 1 )}{4} + \frac{c _{1} ( 1 + t )}{t}

求解步骤

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

求解线性方程:

y = \frac{t ^{3} ( t + 1 )}{4} + \frac{c _{1} ( 1 + t )}{t}

y = \frac{t ^{3} ( t + 1 )}{4} + \frac{c _{1} ( 1 + t )}{t}

\frac{\partial}{\partial x} ((13 x - 1) (3 y - 1))

d er i v a t i v e y = ln (3 x - 7)

\int \frac{2 x ^{3} - x ^{2} + 2 x + 2}{x ^{2} + 1} d x

t a y l or \frac{1}{x - 1}, 10

y^{^{''}} + π^{2} y = 2