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Find the particular anti-derivative of dy/dx = 3/x + 1/x^2, y(1) =1
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High School Teacher
We will use integration here.
`intdy = int(3/x + 1/x^2)dx `
`intdy = int(3/x + 1x^-2)dx`
`y = 3lnx -1/x + c`
Using the formulas: `int(du/u) = lnu + c` and `intx^ndx = x^(n+1)/(n+1) + c`
Now, we will use the given y(1) = 1.
`1 = 3ln(1) - 1/(1) + c`
`1 = 3(0) - 1 + c`
`1 = 0 - 1 + c`
Add 1 on both sides.
`2 = c`
So, the final answer will be:
y = 3lnx -1/x + 2 or y = (3xlnx +2x - 1)/x
Posted by violy on May 8, 2013 at 12:06 AM (Answer #2)
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