Confused about the plus minus sign
#calculus help
junior wind
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Confused about the plus minus sign
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junior wind
the last step before determining the answer is y^-2 = 2/7x^-3 + cx^4
raise both sides to the negative 1/2 power to get y
there are plus/minus signs in the answer choices, how do i go about this?
pliant jay
the last step before determining the answer is y^-2 = 2/7x^-3 + cx^4
Can you show your process?
pliant jay
Ah, I see.
Suppose y^(-2) = x. What is y?
junior wind
1/x^2
pliant jay
What I mean is y^(-2) = x is equivalent to y = ±1/√(x), not just 1/√(x).
That's where ± comes from.
pliant jay
Yeah, seems good.
Though... I think we might be able to solve this a little quicker.
Your approach is good, but I think we can do it with a trick.
junior wind
ok whats the trick?
pliant jay
y' + 2y/x = y^3/x^4
Let's multiply everything by x^2.
x^2 y' + 2xy = y^3/x^2
We notice that x^2 y' + 2xy = (x^2 y)'. So:
(x^2 y)' = y^3/x^2
We want the right part to also contain x^2 y. So, let's multiply its numerator and denominator by x^6.
(x^2 y)' = x^6 y^3/x^8
(x^2 y)' = (x^2 y)^3/x^8
And now this is a separable equation in terms of x^2 y and x.
In general, y' + ay/x = x^(-a)(x^a y)' for any a. That's where the trick comes from.
junior wind
wow that is brilliant
pliant jay
Yeah, this is a pretty useful trick.
Another one that is occasionally useful is:
y' + ay = e^(-ax) (e^(ax) y)'
junior wind
these are p cool but idk if ill remember them lol
pliant jay
Well, you don't need to remember them exactly.