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

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

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y^'sqrt(x)=sqrt(y)cos(sqrt(x))

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} + (1 - n^{2}) y = \frac{2 ( - 1 ) ^{n}}{n}

y^{^{'}} + n^{2} y = 0

y^{^{'}} = - (\frac{1}{3}) y^{2}

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

\frac{d y}{d x} y = x^{2} - 4 x

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