yy^'=2xy+2x

Lösung

y y^{^{'}} = 2 x y + 2 x

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

\frac{d y}{d x} = - 4 e^{x + y}

y^{2} - x^{2} y^{^{'}} = 0

\frac{d y}{d x} = \frac{1 - y ^{2}}{1 + x ^{2}}

y^{^{'}} + (\frac{2 x}{x ^{2} - 1}) y = \frac{x}{- 1 + x ^{2}}

x^{4} \frac{d y}{d x} + y^{5} = 0