- #1

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[tex]y'=x\sqrt{y}[/tex]

like this:

[tex]y^{-\frac{1}{2}}y'=x[/tex]

[tex]\int{y^{-\frac{1}{2}}}dy=\int{xdx}[/tex]

[tex]y^{\frac{1}{2}}=\frac{x^2 +C}{4}[/tex]

[tex]y=\left(\frac{x^2+C}{4}\right)^2[/tex]

Is this the right way to solve it? Because the answer in my textbook says that the answer is

[tex]y=\pm\sqrt{x^2+C}[/tex]

But I really can't see where I've gone wrong.