What is the solution for sqrt(y)dx+(2+x)dy=0,y(-1)=4 ?

The solution for sqrt(y)dx+(2+x)dy=0,y(-1)=4 is y=(ln^2(2+x))/4-(c_{1}ln(x+2))/2+(c_{1}^2)/4

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sqrt(y)dx+(2+x)dy=0,y(-1)=4

y d x + (2 + x) d y = 0, y (- 1) = 4

y = \frac{ln ^{2} ( 2 + x )}{4} - \frac{c _{1} ln ( x + 2 )}{2} + \frac{c _{1}^{2}}{4}

Solution steps

y d x + (2 + x) d y = 0

Solve separable ODE:

y = \frac{ln ^{2} ( 2 + x )}{4} - \frac{c _{1} ln ( x + 2 )}{2} + \frac{c _{1}^{2}}{4}

y = \frac{ln ^{2} ( 2 + x )}{4} - \frac{c _{1} ln ( x + 2 )}{2} + \frac{c _{1}^{2}}{4}

y^{^{'}} = x y + x - y - 1

y^{^{'}} + \frac{2 x}{1 + x ^{2}} y = \frac{- x}{1 + x ^{2}}

y^{^{'}} = (x y)^{2}

\frac{d y}{d x} = 10 x 25 - y^{2}

y^{^{'}} = \frac{13 x}{y + x ^{2} y}

What is the solution for sqrt(y)dx+(2+x)dy=0,y(-1)=4 ?
The solution for sqrt(y)dx+(2+x)dy=0,y(-1)=4 is y=(ln^2(2+x))/4-(c_{1}ln(x+2))/2+(c_{1}^2)/4