yy^'=x-1

Lösung

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

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

Schritte zur Lösung

y y^{'} (x) = x - 1

Solve separable ODE:

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

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

\frac{d y}{d x} = 25 + y^{2}

\frac{d y}{d x} = 19 x + y - 19

\frac{d}{d s} (ln (s))

\int \frac{ln ( 3 x + 1 )}{( 6 x + 2 )} d x

y^{^{'}} = y (3 - t y)