What is the solution for y^'=(1+2*x)sqrt(y) ?

The solution for y^'=(1+2*x)sqrt(y) is y=(c_{1}x^2)/2+(x^3)/2+(x^4)/4+(c_{1}x)/2+(x^2)/4+(c_{1}^2)/4

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y^'=(1+2*x)sqrt(y)

y^{^{'}} = (1 + 2 \cdot x) y

y = \frac{c _{1} x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{4} + \frac{c _{1} x}{2} + \frac{x ^{2}}{4} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (x) = (1 + 2 \cdot x) y

Solve separable ODE:

y = \frac{c _{1} x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{4} + \frac{c _{1} x}{2} + \frac{x ^{2}}{4} + \frac{c _{1}^{2}}{4}

y = \frac{c _{1} x ^{2}}{2} + \frac{x ^{3}}{2} + \frac{x ^{4}}{4} + \frac{c _{1} x}{2} + \frac{x ^{2}}{4} + \frac{c _{1}^{2}}{4}

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

e^{2} x^{^{'}} + x e^{2} + 6 x x^{^{'}} = 0

\frac{d y}{d x} = 4 y^{- 2} (x^{6} + x^{4} + x^{2})

(A + B) d y = C (\frac{x}{y}) d x

\frac{d y}{d x} = \frac{10 + 12 x}{x y ^{2}}

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