What is a particular solution to the differential equation dy/dx=(4sqrtylnx)/x with y(e)=1?

Feb 11, 2017

$y = {\ln}^{4} x$

Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \sqrt{y} \ln x}{x}$

Which is a first order linear separable Differential Equation, so we can rearrange to get:

$\frac{1}{\sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 \ln x}{x}$

and separate the variables to get:

$\int \setminus {y}^{- \frac{1}{2}} \setminus \mathrm{dy} = \int \setminus \frac{4 \ln x}{x} \setminus \mathrm{dx}$

And then we can integrate to get:

${y}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = \left(4\right) \left({\ln}^{2} \frac{x}{2}\right) + C$
$\therefore 2 \sqrt{y} = 2 {\ln}^{2} x + C$

Using $y \left(e\right) = 1$ we get:

$\therefore 2 = 2 {\ln}^{2} e + C$
$\therefore C = 0$

Hence the particular solution is:

$\setminus \setminus \setminus 2 \sqrt{y} = 2 {\ln}^{2} x$
$\therefore \sqrt{y} = {\ln}^{2} x$
$\therefore \setminus \setminus \setminus y = {\ln}^{4} x$

Validation:
1. $x = e \implies y = {\ln}^{4} e = 1 \setminus$ QED
2. $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {\ln}^{3} x}{x} = \frac{4 \ln x}{x} \cdot {\ln}^{2} x = \frac{4 \sqrt{y} \ln x}{x} \setminus$ QED