# First-Order Separable Equation

• June 7th 2011, 04:56 AM
Turlock
First-Order Separable Equation
$\frac{dy}{dx }=\frac{2}{27 }(x-3) \sqrt{\frac{x^2-6x+23}{y } }$

From an earlier part of the question I know that I need to use $f(x)=(x^2 -6x+23)^\frac{3}{2 }$ $f'(x)=\frac{3}{2 }(x^2 -6x+23)^\frac{1}{2 } (2x-6)$

I need a general solution in implicit form and then with y=2 and x=1 an explicit solution
• June 7th 2011, 05:09 AM
Prove It
\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{2}{27}(x - 3)\sqrt{\frac{x^2 - 6x + 23}{y}} \\ \frac{dy}{dx} &= \frac{2(x - 3)\sqrt{x^2 - 6x + 23}}{27\sqrt{y}} \\ \sqrt{y}\,\frac{dy}{dx} &= \frac{2(x - 3)\sqrt{x^2 - 6x + 23}}{27} \end{align*}

You can now integrate both sides with respect to $\displaystyle x$.