xy^'+y=y^2,y(1)=-15

Lösung

x y^{^{'}} + y = y^{2}, y (1) = - 15

y = \frac{15}{- 16 x + 15}

Schritte zur Lösung

x y^{'} (x) + y = y^{2}

Solve Bernoulli ODE:

y = \frac{15}{- 16 x + 15}

y = \frac{15}{- 16 x + 15}

a re a y = 28 - x^{2}, y = 3 x, y = \frac{3}{5} x

\int (\frac{16}{2} x^{3} + 12 x^{5} - 2 x + 1) d x

d er i v a t i v e f (x) = 5 e^{x} (cos (x) - sin (x))

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{0.5})

y^{^{''}} = 1