y^'=1+4y^2

Lösung

y^{^{'}} = 1 + 4 y^{2}

y = \frac{tan ( 2 t + c _{1} )}{2}

Schritte zur Lösung

y^{'} (t) = 1 + 4 y^{2}

Solve separable ODE:

y = \frac{tan ( 2 t + c _{1} )}{2}

y = \frac{tan ( 2 t + c _{1} )}{2}

t an g e n t f (x) = \frac{6 x}{x + 5}, at x = - 3

\frac{d}{d t} (3 t^{3} + s + 3 s^{3} + t)

\int (1 + x + x^{2} + x^{3}) d x

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n = 1 \sum \infty \frac{1}{1 + n}