y^'=(4ty)/(2+t^2),y(0)=4

解答

y^{^{'}} = \frac{4 t y}{2 + t ^{2}}, y (0) = 4

y = (t^{2} + 2)^{2}

求解步骤

y^{'} (t) = \frac{4 t y}{2 + t ^{2}}

求解可分离方程:

y = (t^{2} + 2)^{2}

y = (t^{2} + 2)^{2}

e^{y} y^{^{'}} - x^{2} = 0, y (0) = 0

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

y^{^{'}} - (tan (x)) y = 0

y^{^{'}} + 4 x y - \frac{8 x}{y ^{3}} = 0

\frac{d y}{d x} = \frac{3 x - 5}{y - 2}