Help I don't understand about equailities of the eqation.

  • Thread starter john.lee
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In summary, the question involves finding y when y'=-5xy. The solution involves solving for dy/dx and integrating to find y=+-K*e^(-2.5x^2). However, the answer is given as y=K*e^(-2.5x^2), which does not include the absolute value. This is because the absolute value is used to indicate that y must be positive for the equation to work. In this case, since the equation involves an even power, the +/- can be ignored and the answer is simply y=K*e^(-2.5x^2).
  • #1
john.lee
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Homework Statement



y'=-5xy, y=?



The Attempt at a Solution



i solved that,

dy/dx=-5xy

dy/y=-5x dx

∫(1/y)dy=∫-5x dx

= ln(lyl)= -2.5x^2+C

so, y=+- e^(-2.5x^2+C)
=+-K*e^(-2.5x^2)


but the answer is y=K*e^(-2.5x^2)

how can i understand this?

just shoud i ignore "the absolute value"?
 
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  • #2
john.lee said:

Homework Statement



y'=-5xy, y=?



The Attempt at a Solution



i solved that,

dy/dx=-5xy

dy/y=-5x dx

∫(1/y)dy=∫-5x dx

= ln(lyl)= -2.5x^2+C

so, y=+- e^(-2.5x^2+C)
=+-K*e^(-2.5x^2)


but the answer is y=K*e^(-2.5x^2)

how can i understand this?

just shoud i ignore "the absolute value"?

In your solution ##K = \pm e^C## which can be anything but zero. In the answer ##K## is unrestricted. So the only difference is that the answer includes ##y=0## and your solution doesn't. (It was missed when you divided by ##y##). The answers are the same once you include ##y=0## in yours.
 
  • #3
john.lee said:
just shoud i ignore "the absolute value"?

Yep.

The only reason |abs| is used is to mark the fact that it's impossible to have a ln() value where the thing in the brackets is negative.

Essentially y must be positive for the equation to work.

The only time I can think of that you really need to use +/- is when you have a value with an even power, such as x2, x4, x-6, etc. Because then x can be positive or negative and will still yield the same result when you raise it to that power.

Hope that helps a little!
 

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