Find the solution that solves the initial condition.

**gingerbailey** · September 12th 2008, 06:39 PM

Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition

**mr fantastic** · September 12th 2008, 07:13 PM

Originally Posted by gingerbailey

Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition

The DE can be written as $\frac{dy}{dx} = \frac{\ln x^2}{2x} = \frac{\ln |x| }{x} \Rightarrow y = \int \frac{\ln |x|}{x} \, dx$ .

To solve the integral make the substitution $u = \ln |x| \Rightarrow \frac{du}{dx} = \frac{1}{|x|}$ .

Note that |x| = x when x > 0 and |x| = -x when x < 0.

**Aryth** · September 12th 2008, 07:22 PM

Well, we have:

$2xy' - \ln{(x^2)} = 0$

This is pretty simple to organize, using basic algebra you get:

$2xy' = \ln{(x^2)}$

$y' = \frac{\ln{(x^2)}}{2x}$

Now, we integrate both sides:

$\int y' ~dx = \int \frac{\ln{(x^2)}}{2x} ~dx$

Integrating, we get:

$y = \frac{\ln^2{(x^2)}}{8} + C$

We want to find the solution for when y(1) = 0, so we plug and solve:

$y(1) = \frac{\ln^2{(1^2)}}{8} + C = 8$

$= \frac{0}{8} + C = 8$

$C = 8$

So, our solution is:

$y = \frac{\ln^2{(x^2)}}{8} + 8$

And there you go.

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