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December 18, 2014

December 18, 2014

Posted by **Andres** on Tuesday, October 11, 2011 at 8:15am.

- Calc -
**Reiny**, Tuesday, October 11, 2011 at 8:44amif dy1/dx = 3y1

then y1 = ae^(3x) , where a is a constant, a≠ 0

(check: dy1/dx = 3a(e^(3x)) = 3y1 )

if dy2/dx = 8x + 5

then y2 = 4x^2 + 5x + c

the y-intercept of y1 is (0,a), but that is also the y-intercept of y2

so in y2:

a = 0 +0 + c

a = c

when x=2

y1 = a(e^6) , y2 = 16+10+c = c + 26

but a=c, so

a(e^6) = a+26

a(e^6) - a = 26

a(e^6 - 1) = 26

a = 26/(e^6 - 1) , which would be the y-intercept

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