xy^2y^'=x+x^3,y(0)=1

Lösung

x y^{2} y^{^{'}} = x + x^{3}, y (0) = 1

y = 3 3 x + x^{3} + 1

Schritte zur Lösung

x y^{2} y^{'} (x) = x + x^{3}

Solve separable ODE:

y = 3 3 x + x^{3} + 1

y = 3 3 x + x^{3} + 1

y^{^{'}} = \frac{y - 1}{x + 3}

y^{^{'}} = 5 e^{- x - y}

(t^{2} - 22 t + 85) \frac{d y}{d t} = y

(\frac{d y}{d x}) = (- \frac{1}{( x ^{2} y ^{2} )}), y = 2, x = 1

y^{^{'}} + \frac{1}{3} y = 0