What is the solution for y^'=(1+4x)sqrt(y),y(0)=1 ?

The solution for y^'=(1+4x)sqrt(y),y(0)=1 is y=x^3+x^4+2x^2+(x^2)/4+x+1

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y^'=(1+4x)sqrt(y),y(0)=1

y^{^{'}} = (1 + 4 x) y, y (0) = 1

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

Solution steps

y^{'} (x) = (1 + 4 x) y

Solve separable ODE:

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

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

(x^{- 2} y^{- 5}) d x + (x^{- 3} y^{- 4}) d y = 0

\frac{d V}{d t} = 20 - \frac{8}{100} V

\frac{d y}{d t} = \frac{2 t}{1 + 2 y}

\frac{d y}{d x} + y tan (x) = 0

y 1 - x^{2} \cdot y^{^{'}} + 1 - y^{2} = 0

What is the solution for y^'=(1+4x)sqrt(y),y(0)=1 ?
The solution for y^'=(1+4x)sqrt(y),y(0)=1 is y=x^3+x^4+2x^2+(x^2)/4+x+1