4y^'=1+x+y+xy

Lösung

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

y = e^{\frac{x ^{2}}{8} + \frac{x + c _{1}}{4}} - 1

Schritte zur Lösung

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

Solve separable ODE:

y = e^{\frac{x ^{2}}{8} + \frac{x + c _{1}}{4}} - 1

y = e^{\frac{x ^{2}}{8} + \frac{x + c _{1}}{4}} - 1

y^{^{'}} + 18 y^{^{'}} + 81 y = 0

\frac{d y}{d x} = (- n y)

\frac{d h}{d t} = kh (t)^{n}

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

y + 4 y^{^{'}} + 5 y = 0