y^'=((1+y))/(x^2+x)

Lösung

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

y = \frac{c _{1} x}{x + 1} - 1

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{c _{1} x}{x + 1} - 1

y = \frac{c _{1} x}{x + 1} - 1

in v erse l a pl a ce \frac{1}{3 s}

\frac{\partial}{\partial x} (x^{2} y z^{2})

\frac{\partial}{\partial y} (2 x ln (y) - yz)

s im pl i f y (e^{- x} - e^{5}) (e^{x} + e^{- 2})

y^{^{''}} - 2 y + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0