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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