# Find the solution that solves the initial condition.

• Sep 12th 2008, 06:39 PM
gingerbailey
Find the solution that solves the initial condition.
Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition
• Sep 12th 2008, 07:13 PM
mr fantastic
Quote:

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

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

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

Note that |x| = x when x > 0 and |x| = -x when x < 0.
• Sep 12th 2008, 07:22 PM
Aryth
Well, we have:

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

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

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

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

Now, we integrate both sides:

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

Integrating, we get:

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

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

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

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

$\displaystyle C = 8$

So, our solution is:

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

And there you go.