y^'=(2y+22)^{0.5}

Lösung

y^{^{'}} = (2 y + 22)^{0.5}

y = \frac{2 ^{1 + \frac{1}{2}} t ^{2} + 4 c _{1} t + 2 ^{\frac{1}{2}} c _{1}^{2}}{4 \cdot 2 ^{\frac{1}{2}}} - 11

Schritte zur Lösung

y^{'} (t) = (2 y + 22)^{0.5}

Solve separable ODE:

y = \frac{2 ^{1 + \frac{1}{2}} t ^{2} + 4 c _{1} t + 2 ^{\frac{1}{2}} c _{1}^{2}}{4 \cdot 2 ^{\frac{1}{2}}} - 11

y = \frac{2 ^{1 + \frac{1}{2}} t ^{2} + 4 c _{1} t + 2 ^{\frac{1}{2}} c _{1}^{2}}{4 \cdot 2 ^{\frac{1}{2}}} - 11

d er i v a t i v e f (x) = (x^{3} + 1) \cdot 3 x^{2}

l a pl a ce t r an s f or m t sin (9 t)

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

in v erse l a pl a ce \frac{2 s + 9}{s ^{2} + 4 s + 40}

\frac{\partial}{\partial t} (\frac{t}{c ^{2} t ^{2} - x ^{2}})