Absolute value in solution to separable differential equation

[john.lee]

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"?

 

[LCKurtz]

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.

 

[sa1988]

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!