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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[\int\limits \frac{ dy }{ \sqrt{1-y^2} } = \int\limits 2x dx\] this is as far as iv gotten

  2. IrishBoy123
    • one year ago
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    LHS, use sub: \(y = sin \theta\)

  3. anonymous
    • one year ago
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    see if this will hlp

  4. anonymous
    • one year ago
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    okay \[\int\limits \frac{ dsin }{ 1-\sin^2 } = x^2\]

  5. anonymous
    • one year ago
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    \[\frac{dy}{dx}=2x\sqrt{(1-y^2)}\]\[\frac{dy}{\sqrt{1-y^2}}=2x.dx\] Use the formula \[\int\limits \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C\]

  6. anonymous
    • one year ago
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    okay didn't know that existed , one second

  7. anonymous
    • one year ago
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    \[\sin^{-1}(y) + C\] is this right?

  8. anonymous
    • one year ago
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    y = sin(x^2) + C

  9. anonymous
    • one year ago
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    is that totally wrong?

  10. anonymous
    • one year ago
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    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

  11. anonymous
    • one year ago
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    didn't know any of those ;P anyway what about part B , didn't we just prove their is solution ?

  12. anonymous
    • one year ago
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    \[\sin^{-1}(y)=x^2+C\]\[y(x)=\sin(x^2+C)\] 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

  13. anonymous
    • one year ago
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    okay that actually makes sense, thank you very much!

  14. anonymous
    • one year ago
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    Also, make sure u learn all those formulas, trust me they may look specific but they will help u a ton

  15. anonymous
    • one year ago
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    are those from calc II ? i was told in calc II i would learn better ways of integrating

  16. anonymous
    • one year ago
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    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

  17. anonymous
    • one year ago
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    Have you learned integration by parts and by partial fraction??All these formula's were told us even before those methods

  18. anonymous
    • one year ago
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    woops

  19. anonymous
    • one year ago
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    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?

  20. anonymous
    • one year ago
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    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

  21. anonymous
    • one year ago
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    interesting, what country are you from?

  22. anonymous
    • one year ago
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    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

  23. anonymous
    • one year ago
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    that actually sounds like a cool system i wish i was Indian :D

  24. anonymous
    • one year ago
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    anyway thanks again 0/

  25. anonymous
    • one year ago
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    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

  26. anonymous
    • one year ago
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    still better than having it bee mandatory until 12th grade like it is here

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