1(7+y)y^'=sin(13+x)

Lösung

1 (7 + y) y^{^{'}} = sin (13 + x)

y = - 7 + - 2 (cos (x + 13) - c_{1}) + 49, y = - - 2 (cos (x + 13) - c_{1}) + 49 - 7

Schritte zur Lösung

1 (7 + y) y^{'} (x) = sin (13 + x)

Solve separable ODE:

y = - 7 + - 2 (cos (x + 13) - c_{1}) + 49, y = - - 2 (cos (x + 13) - c_{1}) + 49 - 7

y = - 7 + - 2 (cos (x + 13) - c_{1}) + 49, y = - - 2 (cos (x + 13) - c_{1}) + 49 - 7

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

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

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

\frac{d v}{d t} = 0.5 v - 440

y^{3} - y^{^{'}} + 0.5 y = 0