anonymous
  • anonymous
A._ Solve the differential equation y'= 2x(1 - y^2)^(1/2). B._ Explain why the initial value problem y'= 2x(1-y^2)^(1/2) with y(0) = 3 does not have a solution
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits \frac{ dy }{ \sqrt{1-y^2} } = \int\limits 2x dx\] this is as far as iv gotten
IrishBoy123
  • IrishBoy123
LHS, use sub: \(y = sin \theta\)
anonymous
  • anonymous
https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/separable-equations/v/separable-differential-equations-2

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anonymous
  • anonymous
see if this will hlp
anonymous
  • anonymous
okay \[\int\limits \frac{ dsin }{ 1-\sin^2 } = x^2\]
anonymous
  • anonymous
\[\frac{dy}{dx}=2x\sqrt{(1-y^2)}\]\[\frac{dy}{\sqrt{1-y^2}}=2x.dx\] Use the formula \[\int\limits \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C\]
anonymous
  • anonymous
okay didn't know that existed , one second
anonymous
  • anonymous
\[\sin^{-1}(y) + C\] is this right?
anonymous
  • anonymous
y = sin(x^2) + C
anonymous
  • anonymous
is that totally wrong?
anonymous
  • anonymous
Here are some integrals you should remember \[\int\limits \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C\]\[\int\limits \frac{dx}{a^2+x^2}=\frac{1}{a}.\tan^{-1}(\frac{x}{a})+C\] \[\int\limits \frac{dx}{x^2-a^2}=\frac{1}{2a}\log|\frac{x-a}{x+a}|+C\]\[\int\limits \frac{dx}{a^2-x^2}=\frac{1}{2a}\log|\frac{a+x}{a-x}|+C\] \[\int\limits \frac{dx}{\sqrt{x^2\pm a^2}}=\log|x+\sqrt{x^2\pm a^2}|+C \] and yes ur right
anonymous
  • anonymous
didn't know any of those ;P anyway what about part B , didn't we just prove their is solution ?
anonymous
  • anonymous
\[\sin^{-1}(y)=x^2+C\]\[y(x)=\sin(x^2+C)\] At x=0 \[3=\sin(C)\] This is impossible as the sine function's range is in the interval \[[-1,1]\] So there is no value of C for which sin C will be 3
anonymous
  • anonymous
okay that actually makes sense, thank you very much!
anonymous
  • anonymous
Also, make sure u learn all those formulas, trust me they may look specific but they will help u a ton
anonymous
  • anonymous
are those from calc II ? i was told in calc II i would learn better ways of integrating
anonymous
  • anonymous
I don't know what system is over there, but I learned all these formulas in 12th grade, which would be equivalent of last year of high school over there, in fact integral calculus itself is touched in 12th grade over here
anonymous
  • anonymous
Have you learned integration by parts and by partial fraction??All these formula's were told us even before those methods
anonymous
  • anonymous
woops
anonymous
  • anonymous
no i haven't learned it yet but i will next year ( my last year) im two years ahead in math though. is calculus a requirement in your school?
anonymous
  • anonymous
It depends, if you take maths in 11th grade, then you will have to inevitably study it in 12th as well and thus calculus too
anonymous
  • anonymous
interesting, what country are you from?
anonymous
  • anonymous
If a student opts for arts and humanities he won't have to study maths, but it's compulsory for science students, also if a student opts for commerce he can either have commerce with maths or without maths I'm from India
anonymous
  • anonymous
that actually sounds like a cool system i wish i was Indian :D
anonymous
  • anonymous
anyway thanks again 0/
anonymous
  • anonymous
Haha, not really, all students have to study maths upto 10th grade, this is really bad, some people are just not interested in maths and upto 8th grade mathematics is sufficient for a person to know about, they should make it optional after 8th
anonymous
  • anonymous
still better than having it bee mandatory until 12th grade like it is here

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