(1+x^2)y^'=xy+x^2y^2

Lösung

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

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

Schritte zur Lösung

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

Solve Bernoulli ODE:

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

y = \frac{2 x ^{2} + 1}{- x x ^{2} + 1 + ln ( x ^{2} + 1 + x ) + c _{1}}

t an g e n t f (x) = \frac{1}{7 x}, at x = 9

y^{^{''}} - 2 x y^{^{'}} = 0

\int_{- 2}^{62} 2 π (y) (2) d y

\int \frac{2}{x ^{2} - 4 x} d x

l a pl a ce t r an s f or m sin (2 t) \cdot e^{3 t}