xy^'+y=x^{3/2}y^{5/2}ln(x)

Lösung

x y^{^{'}} + y = x^{\frac{3}{2}} y^{\frac{5}{2}} ln (x)

y = \frac{2 \cdot 2 ^{\frac{1}{3}}}{x ( 9 ln ^{4} ( x ) - 24 c _{1} ln ^{2} ( x ) + 16 c _{1}^{2} ) ^{\frac{1}{3}}}

Schritte zur Lösung

x y^{'} (x) + y = x^{\frac{3}{2}} y^{\frac{5}{2}} ln (x)

Solve Bernoulli ODE:

y = \frac{2 \cdot 2 ^{\frac{1}{3}}}{x ( 9 ln ^{4} ( x ) - 24 c _{1} ln ^{2} ( x ) + 16 c _{1}^{2} ) ^{\frac{1}{3}}}

y = \frac{2 \cdot 2 ^{\frac{1}{3}}}{x ( 9 ln ^{4} ( x ) - 24 c _{1} ln ^{2} ( x ) + 16 c _{1}^{2} ) ^{\frac{1}{3}}}

t an g e n t e^{6 x}

d er i v a t i v e \frac{x + 9}{x}

\frac{d}{d t} (e^{- t} sin (2 t))

s l o p e in t erce pt (- 4, 0), (1, - 5)

t an g e n t f (x) = \frac{1}{2} x^{2}, at x = - 1