Explanation of Absolute Value "Rule" and Manipulating C

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BryanR

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Hi all,

I'm having difficulties understanding what rule, or line of thinking, was employed in step 7 of the problem below. I'm recalling this problem from memory so the original problem/steps might've looked a bit different.

  1. [MATH]\frac{dy}{dx} = x(y+2)[/MATH]
  2. [MATH](y+2)^{-1} dy = x dx[/MATH]
  3. [MATH]\int (y+2)^{-1} dy = \int x dx[/MATH]
  4. [MATH]\ln{|y+2|} = \frac{x^2}{2}+C[/MATH]
  5. [MATH]e^{\ln{|y+2|}} = e^{\frac{x^2}{2}+C}[/MATH]
  6. [MATH]|y+2| = e^{\frac{x^2}{2}}e^C[/MATH]
  7. [MATH]y+2 = Ce^{\frac{x^2}{2}}[/MATH]
  8. [MATH]y = Ce^{\frac{x^2}{2}}-2[/MATH]
I have two questions here. 1) How did they just get rid of the absolute value? 2) Does [MATH]e^C[/MATH] turn into [MATH]C[/MATH] in front of the expression because it's just some other constant? They might've made a distinction between the C in step 6 and the C in steps 7 & 8—I don't quite remember. Thanks.
 
BryanR said:
Hi all,

I'm having difficulties understanding what rule, or line of thinking, was employed in step 7 of the problem below. I'm recalling this problem from memory so the original problem/steps might've looked a bit different.

  1. [MATH]\frac{dy}{dx} = x(y+2)[/MATH]
  2. [MATH](y+2)^{-1} dy = x dx[/MATH]
  3. [MATH]\int (y+2)^{-1} dy = \int x dx[/MATH]
  4. [MATH]\ln{|y+2|} = \frac{x^2}{2}+C[/MATH]
  5. [MATH]e^{\ln{|y+2|}} = e^{\frac{x^2}{2}+C}[/MATH]
  6. [MATH]|y+2| = e^{\frac{x^2}{2}}e^C[/MATH]
  7. [MATH]y+2 = Ce^{\frac{x^2}{2}}[/MATH]
  8. [MATH]y = Ce^{\frac{x^2}{2}}-2[/MATH]
I have two questions here. 1) How did they just get rid of the absolute value? 2) Does [MATH]e^C[/MATH] turn into [MATH]C[/MATH] in front of the expression because it's just some other constant? They might've made a distinction between the C in step 6 and the C in steps 7 & 8—I don't quite remember. Thanks.
Click to expand...
1) How did they just get rid of the absolute value?

Because "exponential functions" are strictly positive \(\displaystyle \to \ \ e^{f(x)} > 0\) for every real value of "f(x)". Thus we can safely loose the || symbol with positive symbol (by convention, implied with no symbol).

2) Does eC turn into C in front of the expression because it's just some other constant?

Yes. But expressing those constants with different symbols will be "good" practice (in an exam that could be the difference between grades A and A+)
 
I look at it differently than Subhotosh Khan does (actually we never see eye to eye)

At step 4 you had [MATH]\ln{|y+2|} = \frac{x^2}{2}+C[/MATH]This says to me that [MATH]|y+2| = e^{\frac{x^2}{2}+C}[/MATH]
Now [MATH]y+2 = \pm e^{\frac{x^2}{2}+C}[/MATH]Note that [MATH] e^{\frac{x^2}{2}+C}>0[/MATH]Note that [MATH] e^{\frac{x^2}{2}+C}[/MATH] can be written as [MATH] |C|e^{\frac{x^2}{2}}[/MATH]Now if we remove the absolute value bars from C we have [MATH]y+2 = \pm e^{\frac{x^2}{2}+C} = Ce^{\frac{x^2}{2}}[/MATH]
 
Some people would object to using the same letter, "C", in \(\displaystyle \pm e^{\frac{x^2}{2}+ C}\) and \(\displaystyle Ce^{\frac{x^2}{2}}\). (Subhotosh Kahn, here's your chance to get back at Jomo!)
 
HallsofIvy said:
Some people would object to using the same letter, "C", in \(\displaystyle \pm e^{\frac{x^2}{2}+ C}\) and \(\displaystyle Ce^{\frac{x^2}{2}}\). (Subhotosh Kahn, here's your chance to get back at Jomo!)
Click to expand...
Haven't you heard? Subhotosh and I are buddies! He even is giving me a 20% pay raise!
 
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