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

Lösung

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

y = e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t^{2} + 2 e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t + e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t^{2} + 2 e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t + e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}}

y = e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t^{2} + 2 e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}} t + e^{\frac{t ^{2} + 2 t + 1}{2} - 2 t - 2 + c_{1}}

d x = \frac{x}{1 - x} d t

(1 - y) y^{^{'}} = x^{2}

(p - 1) y^{^{'}} (p) + 3 y (p) = 0

y^{^{'}} - y - 1 = 0

y^{^{'}} = \frac{- x - x y}{y + x y}