What is the solution for sqrt(x)y^'+sqrt(y)=0 ?

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

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sqrt(x)y^'+sqrt(y)=0

x y^{^{'}} + y = 0

y = x - c_{1} x + \frac{c _{1}^{2}}{4}

Solution steps

x y^{'} (x) + y = 0

Solve separable ODE:

y = x - c_{1} x + \frac{c _{1}^{2}}{4}

y = x - c_{1} x + \frac{c _{1}^{2}}{4}

\frac{d x}{d t} = 7 x t^{3}

y^{^{'}} = x \cdot e^{y}

(x^{3} + 7 x^{3} y^{2}) d x + e^{x^{4}} y d y = 0

(1 + x^{2}) \frac{d y}{d x} + 2 x y = - x

\frac{d y}{d x} y = lo g_{2} (x lo g_{5} (x))

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