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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(1y^2)^(1/2) with y(0) = 3 does not have a solution
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(1y^2)^(1/2) with y(0) = 3 does not have a solution

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{ dy }{ \sqrt{1y^2} } = \int\limits 2x dx\] this is as far as iv gotten

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0LHS, use sub: \(y = sin \theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see if this will hlp

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay \[\int\limits \frac{ dsin }{ 1\sin^2 } = x^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}=2x\sqrt{(1y^2)}\]\[\frac{dy}{\sqrt{1y^2}}=2x.dx\] Use the formula \[\int\limits \frac{dx}{\sqrt{a^2x^2}}=\sin^{1}(\frac{x}{a})+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay didn't know that existed , one second

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin^{1}(y) + C\] is this right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that totally wrong?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here are some integrals you should remember \[\int\limits \frac{dx}{\sqrt{a^2x^2}}=\sin^{1}(\frac{x}{a})+C\]\[\int\limits \frac{dx}{a^2+x^2}=\frac{1}{a}.\tan^{1}(\frac{x}{a})+C\] \[\int\limits \frac{dx}{x^2a^2}=\frac{1}{2a}\log\frac{xa}{x+a}+C\]\[\int\limits \frac{dx}{a^2x^2}=\frac{1}{2a}\log\frac{a+x}{ax}+C\] \[\int\limits \frac{dx}{\sqrt{x^2\pm a^2}}=\logx+\sqrt{x^2\pm a^2}+C \] and yes ur right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0didn't know any of those ;P anyway what about part B , didn't we just prove their is solution ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin^{1}(y)=x^2+C\]\[y(x)=\sin(x^2+C)\] 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
 one year ago
Best ResponseYou've already chosen the best response.0okay that actually makes sense, thank you very much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, make sure u learn all those formulas, trust me they may look specific but they will help u a ton

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are those from calc II ? i was told in calc II i would learn better ways of integrating

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Have you learned integration by parts and by partial fraction??All these formula's were told us even before those methods

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no 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
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0interesting, what country are you from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0that actually sounds like a cool system i wish i was Indian :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyway thanks again 0/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, 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
 one year ago
Best ResponseYou've already chosen the best response.0still better than having it bee mandatory until 12th grade like it is here
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