y^'y^{''}=2

Lösung

y^{^{'}} y^{^{''}} = 2

y = \frac{2}{3} t 4 t + c_{1} + \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}, y = - \frac{2}{3} t 4 t + c_{1} - \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}

Schritte zur Lösung

y^{'} (t) y^{''} (t) = 2

Solve using substitution:

y = \frac{2}{3} t 4 t + c_{1} + \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}, y = - \frac{2}{3} t 4 t + c_{1} - \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}

y = \frac{2}{3} t 4 t + c_{1} + \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}, y = - \frac{2}{3} t 4 t + c_{1} - \frac{1}{6} c_{1} 4 t + c_{1} + c_{2}

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

\frac{d}{d x} ((f (x) (3 x))^{2} + 5)

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

\frac{d}{d x} (ln (x 4 x - 1))

x \to + 0 lim (\frac{sin ^{2} ( x )}{x})