xy^'=(x+1)y

Lösung

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

y = e^{x + c_{1}} x

Schritte zur Lösung

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

Solve separable ODE:

y = e^{x + c_{1}} x

y = e^{x + c_{1}} x

\frac{d y}{d x} + y = 1, y (0) = 0

y^{^{'}} = 15 y - y^{2} - 50

\frac{x ^{2}}{y ^{2} - 3} \frac{d y}{d x} = \frac{1}{2 y}, y (1) = 2

x^{^{'}} = e^{t + x}

d y = x (2 y d x - x d y)