anonymous
  • anonymous
a. Solve the differential equation y prime equals the product of 4 times x and the square root of the quantity 1 minus y squared. b. Explain why the initial value problem y prime equals the product of 4 times x and the square root of the quantity 1 minus y squared with y(0) = 4 does not have a solution.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
https://gyazo.com/fa06c87f78388c31263201846039dc26
terenzreignz
  • terenzreignz
So... go ahead, show me you aren't clueless XD
anonymous
  • anonymous
Lol

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More answers

anonymous
  • anonymous
Not clueless but somewhat stuck
anonymous
  • anonymous
I remember this but still quite lost
anonymous
  • anonymous
How would I solve
anonymous
  • anonymous
\[1-y^2=e^{2x^2+c}\]
anonymous
  • anonymous
For C?
anonymous
  • anonymous
Need a refreshing mint
terenzreignz
  • terenzreignz
What in the world... haha To be honest, I haven't even started trying to solve it myself :> Hang on...
anonymous
  • anonymous
Haha
anonymous
  • anonymous
There's a way for sure, just can't seem to remember
terenzreignz
  • terenzreignz
\[\Large \frac{dy}{dx}=4x\sqrt{1-y^2}\]
terenzreignz
  • terenzreignz
\[\Large \frac {dy}{\sqrt{1-y^2}}=4xdx\] You got this far?
anonymous
  • anonymous
Yup and integrated.
terenzreignz
  • terenzreignz
And you got...?
anonymous
  • anonymous
\[\ln(\sqrt{1-y^2})=2x^2+C\]
terenzreignz
  • terenzreignz
Ahh. No.
terenzreignz
  • terenzreignz
Not even close :P
terenzreignz
  • terenzreignz
Humour me and try differentiating \[\Large \ln \sqrt{1-y^2}\] You'll see you won't get to the original \[\Large \frac1{\sqrt{1-y^2}}\]
anonymous
  • anonymous
Is that not the original?
anonymous
  • anonymous
\[\frac{ y }{ y^2-1 }\]
terenzreignz
  • terenzreignz
No. Chain rule.
terenzreignz
  • terenzreignz
And the original is \[\Large \frac1{\sqrt{1-y^2}}\]
anonymous
  • anonymous
Oh
anonymous
  • anonymous
It is a property of inverse sin
anonymous
  • anonymous
Am I right or right?
terenzreignz
  • terenzreignz
Finally ^^ \[\Large \sin^{-1}(y) = 2x^2 + C\]
anonymous
  • anonymous
Now what?
terenzreignz
  • terenzreignz
Well, the second part of the question. Show that the initial value y(0) = 4 has no solution.
terenzreignz
  • terenzreignz
Should be easy enough?
anonymous
  • anonymous
So wait....that's the end of the first part? No way to make it y=?
anonymous
  • anonymous
I honestly don't think so but sometimes you poop magic out
anonymous
  • anonymous
Magic that makes sense.
terenzreignz
  • terenzreignz
Well, you can certainly have \[\Large y = \sin[2x^2 + C]\]
terenzreignz
  • terenzreignz
It doesn't really matter. Now, do the second part. :D
anonymous
  • anonymous
I like that better. Second part it is
anonymous
  • anonymous
You get
anonymous
  • anonymous
\[4=\sin(c)\]
anonymous
  • anonymous
Which can't be solved.
terenzreignz
  • terenzreignz
Sehr gut ^^ Very good.
anonymous
  • anonymous
wut?
terenzreignz
  • terenzreignz
Sehr gut is "very good" in German :D
anonymous
  • anonymous
TIL
terenzreignz
  • terenzreignz
Are we done here? :D

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