(1+x)y^'=4y

Lösung

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

y = e^{c_{1}} + 4 e^{c_{1}} x + 6 e^{c_{1}} x^{2} + 4 e^{c_{1}} x^{3} + e^{c_{1}} x^{4}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{c_{1}} + 4 e^{c_{1}} x + 6 e^{c_{1}} x^{2} + 4 e^{c_{1}} x^{3} + e^{c_{1}} x^{4}

y = e^{c_{1}} + 4 e^{c_{1}} x + 6 e^{c_{1}} x^{2} + 4 e^{c_{1}} x^{3} + e^{c_{1}} x^{4}

x d y + 1 + y^{2} d x = 0

\frac{d y}{d x} + x y = \frac{x}{y ^{2}}, y (1) = 0.5

\frac{1}{x} \frac{d y}{d x} + 4 y = 2

(3 x + 8) (y^{2} + 4) d x - 4 y (x^{2} + 5 x + 6) d y = 0

\frac{d y}{d x} = \frac{y ln ( x )}{x}