y^'+t/(y(t^2-1))=0

解答

y^{^{'}} + \frac{t}{y ( t ^{2} - 1 )} = 0

y = - ln (t^{2} - 1) + c_{1}, y = - - ln (t^{2} - 1) + c_{1}

求解步骤

y^{'} (t) + \frac{t}{y ( t ^{2} - 1 )} = 0

求解可分离方程:

y = - ln (t^{2} - 1) + c_{1}, y = - - ln (t^{2} - 1) + c_{1}

y = - ln (t^{2} - 1) + c_{1}, y = - - ln (t^{2} - 1) + c_{1}

y^{^{'}} = e^{9 x - 9 y}

y^{^{'}} = \frac{e ^{y}}{cos ^{2} ( t )}

\frac{d y}{d x} = a y + b

\frac{1}{x} d x = \frac{1}{y} d y

x y^{2} y^{^{'}} = ln (x)