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# seperating xy by integration watch

1. ix) dy/dx=xe^y
find the general solution by seperating the variables

integrate 1/e^y dy = integrate x dx
-e^-y=(x^2/2)+c

log it:
-y= - ln (1/2x^2 +c)

but the answer says it is
y= -ln(c-(1/2x^2))

I'm not sure how it got c-.....
2. (Original post by 0utdoorz)
ix) dy/dx=xe^y
find the general solution by seperating the variables

integrate 1/e^y dy = integrate x dx
-e^-y=(x^2/2)+c

log it:
-y= - ln (1/2x^2 +c)

but the answer says it is
y= -ln(c-(1/2x^2))

I'm not sure how it got c-.....
c is just a constant so they relabelled it -c.
3. (Original post by davros)
c is just a constant so they relabelled it -c.
so can it also be +c?
4. You've made a sign error. Make the exponential term positive, then take the log.
-e^(-y)=x^2/2 + c
e^(-y)=-x^2/2 + c

The problem is log(-e^-y)=/= -y Remember the log function is only defined for positive values.
5. (Original post by 0utdoorz)
ix) dy/dx=xe^y
find the general solution by seperating the variables

integrate 1/e^y dy = integrate x dx
-e^-y=(x^2/2)+c

log it:
-y= - ln (1/2x^2 +c)

but the answer says it is
y= -ln(c-(1/2x^2))

I'm not sure how it got c-.....
What I'd say is you have:

However your next steps seems faulty. You don't want to be taking the log of a negative value, so:

as c can be positive or negative as it is a constant.

Then takes logs:

so:

as you want.

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