What is the solution for y^'=(2y+22)^{0.5} ?

The solution for y^'=(2y+22)^{0.5} is y=(2^{1+1/2}t^2+4c_{1}t+2^{1/2}c_{1}^2)/(4*2^{1/2)}-11

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

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

Solution steps

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

f (x) = (x^{3} + 1) \cdot 3 x^{2}

t sin (9 t)

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

\frac{2 s + 9}{s ^{2} + 4 s + 40}

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

What is the solution for y^'=(2y+22)^{0.5} ?
The solution for y^'=(2y+22)^{0.5} is y=(2^{1+1/2}t^2+4c_{1}t+2^{1/2}c_{1}^2)/(4*2^{1/2)}-11