What is the solution for 4xy^'-3=y^2+2 ?

The solution for 4xy^'-3=y^2+2 is y=tan((sqrt(5)(ln(x)+c_{1}))/4)sqrt(5)

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4xy^'-3=y^2+2

4 x y^{^{'}} - 3 = y^{2} + 2

y = tan (\frac{5 ( ln ( x ) + c _{1} )}{4}) 5

Solution steps

4 x y^{'} (x) - 3 = y^{2} + 2

Solve separable ODE:

y = tan (\frac{5 ( ln ( x ) + c _{1} )}{4}) 5

y = tan (\frac{5 ( ln ( x ) + c _{1} )}{4}) 5

(1 + x) \frac{d y}{d x} = e^{- y} - 1

\frac{1500}{230} (1 - 30 x^{^{'}}) = \frac{x}{0.1} + 75 x^{^{'}}

y^{^{'}} + a y + \frac{2 \cdot y}{x} = 0

\frac{d P}{d t} = P - P^{- 2}

y^{^{'}} = \frac{2 y}{7 x}

What is the solution for 4xy^'-3=y^2+2 ?
The solution for 4xy^'-3=y^2+2 is y=tan((sqrt(5)(ln(x)+c_{1}))/4)sqrt(5)