(x^2+1)y^'+8x(y-1)=0,y(0)=3

Lösung

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 3

y = \frac{2}{( x ^{2} + 1 ) ^{4}} + 1

Schritte zur Lösung

(x^{2} + 1) y^{'} (x) + 8 x (y - 1) = 0

Solve separable ODE:

y = \frac{2}{( x ^{2} + 1 ) ^{4}} + 1

y = \frac{2}{( x ^{2} + 1 ) ^{4}} + 1

\frac{\partial}{\partial x} (4 x^{2} - 4 x y + x yz)

\int (t^{2} - 1) (4 + 3 t) d t

t an g e n t f (x) = x^{4}, at x = 1

t an g e n t y = \frac{6}{x}, (25, \frac{6}{5})

x \to \infty lim (\frac{( 2 x ^{2} + 1 )}{3 x - 5})