# How to you find the general solution of sqrt(x^2-9)y'=5x?

Jan 18, 2017

$y = 5 \sqrt{{x}^{2} - 9} + C$

#### Explanation:

Writing $y '$ as $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\sqrt{{x}^{2} - 9} \frac{\mathrm{dy}}{\mathrm{dx}} = 5 x$

Then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 x}{\sqrt{{x}^{2} - 9}}$

Separating the variables by treating $\frac{\mathrm{dy}}{\mathrm{dx}}$ as a quotient then integrating both sides:

$\int \mathrm{dy} = \int \frac{5 x}{\sqrt{{x}^{2} - 9}} \mathrm{dx}$

$y = \int \frac{5 x}{\sqrt{{x}^{2} - 9}} \mathrm{dx}$

Solve the remaining integral by letting $u = {x}^{2} - 9$, implying that $\mathrm{du} = 2 x \textcolor{w h i t e}{.} \mathrm{dx}$:

$y = \frac{5}{2} \int \frac{2 x}{\sqrt{{x}^{2} - 9}} \mathrm{dx}$

$y = \frac{5}{2} \int \frac{1}{\sqrt{u}} \mathrm{du}$

$y = \frac{5}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$y = \frac{5}{2} \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right) + C$

$y = 5 \sqrt{u} + C$

$y = 5 \sqrt{{x}^{2} - 9} + C$