(x+y)^2+(2xy+x^2-1)(dy)/(dx)=0

Lösung

(x + y)^{2} + (2 x y + x^{2} - 1) \frac{d y}{d x} = 0

y = \frac{3 - 3 x ^{2} + 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{3 - 3 x ^{2} - 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

Schritte zur Lösung

(x + y)^{2} + (2 x y + x^{2} - 1) \frac{d y}{d x} = 0

Solve exact ODE:

y = \frac{3 - 3 x ^{2} + 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{3 - 3 x ^{2} - 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

y = \frac{3 - 3 x ^{2} + 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}, y = \frac{3 - 3 x ^{2} - 9 - 18 x ^{2} - 3 x ^{4} + 36 c _{1} x}{6 x}

\int_{- 2}^{2} x 4 - x^{2} d x

a re a y = 4 x^{2}, y = x^{2} + 3

\int_{- 1}^{4} (6 x - 9) d x

x \to \infty lim (\frac{x}{n})

\int_{3}^{8} \frac{1}{x ^{2} - 4} d x