y^'=1+y

Lösung

y^{^{'}} = 1 + y

y = e^{t + c_{1}} - 1

Schritte zur Lösung

y^{'} (t) = 1 + y

Solve separable ODE:

y = e^{t + c_{1}} - 1

y = e^{t + c_{1}} - 1

\int_{0}^{1} \frac{1}{x - 1} d x

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

in v erse l a pl a ce \frac{4}{( 3 s - 1 ) ^{5}}

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

\frac{d}{d x} (\frac{4 x + 2}{( 3 - 2 x ) ^{3}})