What is the solution for (dy)/(dx)=((x^2-1))/(2y),y(3)=1 ?

The solution for (dy)/(dx)=((x^2-1))/(2y),y(3)=1 is y=sqrt((x^3)/3-x-5)

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(dy)/(dx)=((x^2-1))/(2y),y(3)=1

\frac{d y}{d x} = \frac{( x ^{2} - 1 )}{2 y}, y (3) = 1

y = \frac{x ^{3}}{3} - x - 5

Solution steps

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

Solve separable ODE:

y = \frac{x ^{3}}{3} - x - 5

y = \frac{x ^{3}}{3} - x - 5

\frac{\partial}{\partial y} (y^{2} e^{x yz})

\frac{\partial}{\partial x} (y^{5} - 4 x y)

y^{^{'}} = \frac{6 y ^{3}}{( 3 - 2 x ) ^{2}}, y (0) = - 1

\int_{0}^{1} ln (2 x) d x

\int (\frac{2}{x ^{2}}) d x

What is the solution for (dy)/(dx)=((x^2-1))/(2y),y(3)=1 ?
The solution for (dy)/(dx)=((x^2-1))/(2y),y(3)=1 is y=sqrt((x^3)/3-x-5)