-3y^'+(2y)/x =((y^4))/(x^4)

Lösung

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

y = \frac{x}{3 - 1 + c _{1} x}

Schritte zur Lösung

- 3 y^{'} (x) + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

Solve Bernoulli ODE:

y = \frac{x}{3 - 1 + c _{1} x}

y = \frac{x}{3 - 1 + c _{1} x}

\frac{d y}{d x} = \frac{x - y - 3}{x - 3}

in v erse l a pl a ce \frac{1}{0.012 s ^{2} + 1.55 s + 50}

s \to 1 lim (\frac{( 2 s + 3 )}{s})

\frac{d y}{d x} = 2 y + 3

\int_{1}^{8} (x^{2} - 4 x + 6) d x