y^2=(xyy^'+1)ln(x)

Lösung

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

y = 2 ln (x) + c_{1} ln^{2} (x), y = - 2 ln (x) + c_{1} ln^{2} (x)

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = 2 ln (x) + c_{1} ln^{2} (x), y = - 2 ln (x) + c_{1} ln^{2} (x)

y = 2 ln (x) + c_{1} ln^{2} (x), y = - 2 ln (x) + c_{1} ln^{2} (x)

y^{^{'}} = 1 - y

t^{2} y^{^{'}} + 6 t y - y^{3} = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0

d er i v a t i v e e^{x^{2} - 4 x}

d er i v a t i v e 8 e^{4 x}