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

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- Sep 12th 2008, 07:39 PMgingerbaileyFind the solution that solves the initial condition.
Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition

- Sep 12th 2008, 08:13 PMmr fantastic
- Sep 12th 2008, 08:22 PMAryth
Well, we have:

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

Now, we integrate both sides:

Integrating, we get:

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

So, our solution is:

And there you go.