Question

calc help

Answer 1

@jim_thompson5910

Answer 2

This d. e. is separable. Have you considered using that approach to solving the d. e.?

Answer 3

what is d.e?

Answer 4

differential equation

Answer 5

6x dx and \[\sqrt{1-y ^{2}} dy\] ?

Answer 6

\[\sqrt{1-y ^{2}} would be \in the denominator on the \left side, under dy.\]

Answer 7

\[\frac{ dy }{ \sqrt{1-y ^{2}} }\]

Answer 8

Yes, the right side would be 6xdx.

Answer 9

Integrate both sides of this d. e. now.

Answer 10

y-(y^3/3)+c= 3x^2+c

Answer 11

What happened to the radical?

Answer 12

i took the antiderivative and got

y^2dy=x^3/3

Answer 13

y^3* not x^3

Answer 14

Better check that. The correct integral of the left side is an inverse trig function.
Please use a reference book...find out how to integrate the left side. I used a trig substitution.

Answer 15

arcsin(y)+C

Answer 16

@mathmale

Answer 17

@satellite73

Answer 18

@zepdrix @jim_thompson5910

Answer 19

This is what we mean by separable equations. we get our dx and x on the same side as well as y and dy.

\[\frac{ dy }{ dx } = 6x \sqrt{1-y^{2}}\]

\[dx*\frac{ dy }{ dx } = 6x \sqrt{1-y^{2}}*dx\]

\[\frac{ dy }{ \sqrt{1-y^{2}} } = 6x~dx\]

after this I won't go any further.

Answer 20

arcsin(y)+c= 3x^2+c

Answer 21

y=sin(3x^2+c)

Answer 22

I'm not sure what part B is asking

Answer 23

y(0) = 2

Answer 24

I think it's asking for a general solution when you plug in x = 0.

Answer 25

\[\large\rm y=\sin(3x^2+c)\]They want know why this initial data is a problem,\[\large\rm 2=\sin(3\cdot0^2+c)\]Think about it a sec :)
Something bad here.

Answer 26

:) so inverse since of arc sine is sine right lol?

Answer 27

Yes, that step looked ok :D

Answer 28

I can't figure out whats wrong. Is it bc we dont know c

Answer 29

No, it has something to do with sines largest and smallest possible values.

Answer 30

yeah, something bad is happening with that equation :S

Answer 31

2 doesnt equal 0 if you solve

Answer 32

Remember your sine function?
How big can it get?
how small?