What is the solution for y^'= 1/2 xsqrt(1-16y^2),y(0)=0 ?

The solution for y^'= 1/2 xsqrt(1-16y^2),y(0)=0 is y=(sin(x^2))/4

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y^'= 1/2 xsqrt(1-16y^2),y(0)=0

y^{^{'}} = \frac{1}{2} x 1 - 16 y^{2}, y (0) = 0

y = \frac{sin ( x ^{2} )}{4}

Solution steps

y^{'} (x) = \frac{1}{2} x 1 - 16 y^{2}

Solve separable ODE:

y = \frac{sin ( x ^{2} )}{4}

y = \frac{sin ( x ^{2} )}{4}

y^{^{'}} = 4 y^{- 2} (x^{6} + x^{4} + x^{2}), y (0) = 0

\frac{d y}{d x} = - x^{5} y^{2}

\frac{d P}{d t} = 7 Pt, P (1) = 2

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

cos (x) d x + (1 + \frac{1}{y}) sin (x) d y = 0

What is the solution for y^'= 1/2 xsqrt(1-16y^2),y(0)=0 ?
The solution for y^'= 1/2 xsqrt(1-16y^2),y(0)=0 is y=(sin(x^2))/4