2002 STEP 2 Q 8

for the first 2 parts 'find y in terms of x...'

How do you remove the modulus sign on the y after integration e.g. for the first part

1/y dy = -1 dx

ln mod(y) = -x + c

mod(y) = Ae^-x

even after substituting the initial conditions, how can I justify in removing the modulus sign on the y?

y = (+ or -)Ae^-x but because

y = (+ or -)Ae^-x but because A is an arbitrary constant you can just write y = Ae^-x.

Thank you - is this the right

Thank you - is this the right way to justify:
when y>=0 y=Ae^-x
When y<0 -y=Ae^-x y=-Ae^-x

Hence when y is real y= (+ or -) Ae^-x since A is a constant we can say y=Be^-x and then substitute the conditions?

This works

I said in my other post not to stress to much about justification, but this is definitely a reasonable way to justify it if you want to!

A general principle

I used to find this pretty confusing too. The way I look at it now is that to solve a differential equation, you just have to find a function that works - everything up to that point is just a method to help you do that. So you don't really have to justify in any detail why you drop the modulus signs - you can just try y=Ae^-x and see that it works, so we've found the solution.

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