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

Lösung

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

y = \frac{- 16 + 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}, y = \frac{- 16 - 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{- 16 + 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}, y = \frac{- 16 - 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}

y = \frac{- 16 + 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}, y = \frac{- 16 - 256 - 4 ( - c _{1} + 2 cos ( 6 + x ) )}{2}

\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}

(1 + e^{x}) \cdot y \cdot y^{^{'}} = e^{x}, y (0) = 1