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

Lösung

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

y = e^{c_{1}} x^{8} - 4 e^{c_{1}} x^{7} + 10 e^{c_{1}} x^{6} - 16 e^{c_{1}} x^{5} + 19 e^{c_{1}} x^{4} - 16 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{2} - 4 e^{c_{1}} x + e^{c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

y = e^{c_{1}} x^{8} - 4 e^{c_{1}} x^{7} + 10 e^{c_{1}} x^{6} - 16 e^{c_{1}} x^{5} + 19 e^{c_{1}} x^{4} - 16 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{2} - 4 e^{c_{1}} x + e^{c_{1}}

y = e^{c_{1}} x^{8} - 4 e^{c_{1}} x^{7} + 10 e^{c_{1}} x^{6} - 16 e^{c_{1}} x^{5} + 19 e^{c_{1}} x^{4} - 16 e^{c_{1}} x^{3} + 10 e^{c_{1}} x^{2} - 4 e^{c_{1}} x + e^{c_{1}}

x^{2} (x d y + y d x) = 4 x d y - 4 y d x

\frac{d y}{d x} = - 800 y

\frac{d y}{d x} = 14 x y

\frac{2}{x} y^{^{'}} + y = 0

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