y^'=(x-3)(y+1)^{2/3}

Lösung

y^{^{'}} = (x - 3) (y + 1)^{\frac{2}{3}}

y = - x^{3} + \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{c _{1} x ^{4}}{36} + \frac{x ^{4}}{2} + \frac{- c _{1} x ^{3} - c _{1}^{2} x}{3} + \frac{c _{1}^{2} x ^{2}}{18} + c_{1} x^{2} + \frac{c _{1}^{3}}{27} - 1

Schritte zur Lösung

y^{'} (x) = (x - 3) (y + 1)^{\frac{2}{3}}

Solve separable ODE:

y = - x^{3} + \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{c _{1} x ^{4}}{36} + \frac{x ^{4}}{2} + \frac{- c _{1} x ^{3} - c _{1}^{2} x}{3} + \frac{c _{1}^{2} x ^{2}}{18} + c_{1} x^{2} + \frac{c _{1}^{3}}{27} - 1

y = - x^{3} + \frac{x ^{6}}{216} - \frac{x ^{5}}{12} + \frac{c _{1} x ^{4}}{36} + \frac{x ^{4}}{2} + \frac{- c _{1} x ^{3} - c _{1}^{2} x}{3} + \frac{c _{1}^{2} x ^{2}}{18} + c_{1} x^{2} + \frac{c _{1}^{3}}{27} - 1

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

\frac{d y}{d x} = cos (x)

(2 y + 2 t - 7) d t + (4 y + 2 t - 1) d y = 0, y (- 1) = 2

d er i v a t i v e f (x) = 2 x - 3

d er i v a t i v e f (x) = (1 + 3 x^{4})^{2} + 3 x^{3}