18y^'=y(y^2-9)

Lösung

18 y^{^{'}} = y (y^{2} - 9)

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

Schritte zur Lösung

18 y^{'} (t) = y (y^{2} - 9)

Solve separable ODE:

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

y = \frac{3}{1 - e ^{t + c_{1}}}, y = - \frac{3}{1 - e ^{t + c_{1}}}

x \to 1 lim (\frac{1}{( 1.5 - 1 ) ^{2}})

\frac{\partial}{\partial x} (tan (4 x y^{5}))

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a re a x^{2} - 2 x + 3, x = - 2, x = 1

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