Rewrite the differential equation [tex]\frac{dy}{dx}=x{\sqrt{y}}[/tex] in the form y=f(x) given the initial condition f(3)=25.(adsbygoogle = window.adsbygoogle || []).push({});

I am new to integration so I am unsure about my work on this problem.

[tex]\frac{dy}{dx}=x{\sqrt{y}}[/tex]

[tex]dy=(dx)(x)(\sqrt{y})[/tex]

[tex]\frac{dy}{\sqrt{y}}=(dx)(x)[/tex]

[tex]\int{\frac{dy}{\sqrt{y}}}=\int{(x)(dx)}[/tex]

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

[tex]10=\frac{9}{2}+C[/tex]

[tex]C=\frac{11}{2}[/tex]

[tex]2y^{\frac{1}{2}}=\frac{1}{2}x^2+\frac{11}{2}[/tex]

[tex]y=(\frac{1}{4}x^2+\frac{11}{4})^2[/tex]

If I did it correctly, is there an easier way to do it? If I messed up, where?

Thanks

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# Initial value problem

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