Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition
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Originally Posted by gingerbailey Differential Equation: 2xy' - ln (x^2) = 0, y(1)=8 Initial Condition The DE can be written as .
To solve the integral make the substitution .
Note that |x| = x when x > 0 and |x| = -x when x < 0.
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.
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