y^'=4x+xy,y(0)=-1

Lösung

y^{^{'}} = 4 x + x y, y (0) = - 1

y = 3 e^{\frac{x ^{2}}{2}} - 4

Schritte zur Lösung

y^{'} (x) = 4 x + x y

Solve separable ODE:

y = 3 e^{\frac{x ^{2}}{2}} - 4

y = 3 e^{\frac{x ^{2}}{2}} - 4

\frac{d y}{d x} y = 5 x^{3}

\frac{d y}{d x} 2 x + y = 0

y^{^{'}} = - \frac{2}{4} x^{3} y^{8}

y^{^{'}} = - t y^{3}

\frac{d y}{d x} y = (3 + x)^{\frac{5}{x}}