y^'=(t^2+ty+y^2)/(t^2)

Lösung

y^{^{'}} = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}

y = tan (ln (t) + c_{1}) t

Schritte zur Lösung

y^{'} (t) = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}

Solve using substitution:

y = tan (ln (t) + c_{1}) t

y = tan (ln (t) + c_{1}) t

\frac{\partial}{\partial x} (cos^{2} (x + y))

\frac{d y}{d x} + \frac{2 y}{x} = \frac{1}{x ^{2}}

a re a x^{2} - 4, y = 0, x = - 4, x = 3

\frac{d}{d x} (ln (\frac{a + b x}{a - b x}))

in v erse l a pl a ce \frac{1}{s ^{2} + 4 s + 12}