- #1

Titans86

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## Homework Statement

[tex]\frac{dy}{dx} = \sqrt{xy^3} , y(0) = 4[/tex]

## The Attempt at a Solution

So;

[tex]\frac{dy}{dx} = x^{\frac{1}{2}}y^{\frac{3}{2}}

\Rightarrow \int y^{-\frac{3}{2}}dy = \int x^{\frac{1}{2}}dx

\Rightarrow -2y^{-\frac{1}{2}} = \frac{2}{3}x^{\frac{3}{2}} + C

\Rightarrow y^{-\frac{1}{2}} = -\left(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C\right)

\Rightarrow y^{\frac{1}{2}} = -\left(\frac{1}{\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C}\right)

\Rightarrow y = \frac{1}{(\frac{1}{3}x^{\frac{3}{2}} + \frac{1}{2}C)^2}

[/tex]

Then Putting in the conditions mentioned above:

[tex]

4 = \frac{1}{(0 + \frac{1}{2}C)^2}

\Rightarrow \frac{1}{4}C^2 = \frac{1}{4}

\Rightarrow C = 1

[/tex]

Yet my book shows [tex]C = \frac{3}{2}[/tex]