- 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

- schrodinger

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- anonymous

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

- IrishBoy123

LHS, use sub: \(y = sin \theta\)

- anonymous

https://www.khanacademy.org/math/differential-equations/first-order-differential-equations/separable-equations/v/separable-differential-equations-2

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## More answers

- anonymous

see if this will hlp

- anonymous

okay \[\int\limits \frac{ dsin }{ 1-\sin^2 } = x^2\]

- 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

okay didn't know that existed , one second

- anonymous

\[\sin^{-1}(y) + C\]
is this right?

- anonymous

y = sin(x^2) + C

- anonymous

is that totally wrong?

- 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

didn't know any of those ;P
anyway what about part B , didn't we just prove their is solution ?

- 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

okay that actually makes sense, thank you very much!

- anonymous

Also, make sure u learn all those formulas, trust me they may look specific but they will help u a ton

- anonymous

are those from calc II ? i was told in calc II i would learn better ways of integrating

- 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

Have you learned integration by parts and by partial fraction??All these formula's were told us even before those methods

- anonymous

woops

- 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

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

interesting, what country are you from?

- 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

that actually sounds like a cool system i wish i was Indian :D

- anonymous

anyway thanks again 0/

- 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

still better than having it bee mandatory until 12th grade like it is here

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