-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

(8 + y) y^{^{'}} = sin (6 + x)

\frac{d x}{d t} = 14 - 0.56 x, t (0) = 130

d x = \frac{1}{1 + y ^{2}} d y

x - \frac{y}{y ^{^{'}}} = 2 x - 3 x^{2}

y = 3 sin (x) \frac{d y}{d x}