y^'=(y+2t)^2,y(0)=sqrt(3)

Lösung

y^{^{'}} = (y + 2 t)^{2}, y (0) = 3

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

Schritte zur Lösung

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

Solve using substitution:

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

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

n = 1 \sum \infty \frac{4}{( n + 1 ) !}

d er i v a t i v e \frac{5}{4} x^{\frac{4}{3}}