Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

see if this will hlp

Answer 5

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

Answer 6

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

Answer 7

okay didn't know that existed , one second

Answer 8

\[\sin^{-1}(y) + C\]

is this right?

Answer 9

y = sin(x^2) + C

Answer 10

is that totally wrong?

Answer 11

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

Answer 12

didn't know any of those ;P

anyway what about part B , didn't we just prove their is solution ?

Answer 13

\[\sin^{-1}(y)=x^2+C\]\[y(x)=\sin(x^2+C)\]
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

Answer 14

okay that actually makes sense, thank you very much!

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

woops

Answer 20

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?

Answer 21

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

Answer 22

interesting, what country are you from?

Answer 23

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

Answer 24

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

Answer 25

anyway thanks again 0/

Answer 26

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

Answer 27

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