(x^2+1)y^'+4xy=x

Lösung

(x^{2} + 1) y^{^{'}} + 4 x y = x

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

Schritte zur Lösung

(x^{2} + 1) y^{'} (x) + 4 x y = x

Solve separable ODE:

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

y = \frac{c _{1}}{4 ( x ^{4} + 2 x ^{2} + 1 )} + \frac{1}{4}

\frac{d}{d x} (x^{2} y^{2} + x y) = 0

y^{^{'}} = \frac{1 + y}{tan ( x )}

(x^{2} + 1) y^{^{'}} + 4 x y = 0

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

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