Question

Having some problems doing a Differential equation,
So it says dy/dx=(y*cos(x))/(1-y^2) with y(0)=1
I have gotten it past the point of integrating both sides but I am having trouble isolating the y to the left side of the equation, here is my integrated problem: y^2/2-ln(y)=-sin(x)+c ( I originally got ln|y|-y^2/2 for the left but found when checking my work with a basic integration calculator it seems to indicate that the Absolute value signs were not necessary, some concrete explanation on that would be appreciated) But i got it to the point where i have y^2-y=e^(-2sin(x)+2c)

Answer 1

But continuing too much further is seeming to cause my y to cancel, Would really appreciate a little guidence so i am more sure of myself here

Answer 2

you have solved the DE
plug in the initial value and solve for the consntant c

Answer 3

Yes but Should i not isolate y before continuing?

also wanted to make sure everything i did is correct\

Answer 4

many DE cannot be solve to the form y = ...

y^2/2 - ln(y) = ... is fine

Answer 5

Wow really? all this time I have had quite a few i could not isolate i always thought it was error on my part, in that case it would be..... 1^2/2 - ln(1)= -sin(0)+c which gives me a value of 1/2 for c?

Answer 6

therefore y^2/2 -ln(y) = -sin(x) +1/2 ?

Answer 7

yep

Answer 8

what do you get if you take the derivative on both sides of this equation (with respect to x)?

Answer 9

A bit confused by this forgive me for not understanding as I did not do a good job of associating vocabulary with the methodology of this chapter, i find myself doing without knowing the names of the strategies most likely i know how to do it just do not understand the term for it XD
s

Answer 10

Oh hold on

Answer 11

i thought that said what happend if you take the implicate derivative with respect to x

Answer 12

so simply isolating to X after i integrate?

Answer 13

it did, i changed it because the words

Answer 14

take this equation (your solution)

y^2/2 - ln(y) = -sin(x) +1/2

take the derivative on both sides

you should get something like

( f(x) ) dy/dx = g(x)

Answer 15

y-1/y dy/dx=-sin(x)?

been far too long since i did one of these

Answer 16

right, now solve for dy/dx

Answer 17

Correction it should be -cos(x) on the right

Answer 18

oh yeah,

Answer 19

so then (-ycos(x))/y-1=dy/dx

Answer 20

seems liek im close but perhaps my final equation b4 deriving was not correct?

Answer 21

so you have

(y-1/y) dy/dx = -cos(x)
dy/dx = -cos(x)/(y-1/y)
= -ycos(x)/(y^2-1)
= -ycos(x)/(1-y^2)

Answer 22

Having trouble with were that square is coming from

Answer 23

I see you inversed and mult, but..

Answer 24

ok
we had something like
\[\frac{-1}{y-1/y}\]
we multiply by one
\[\frac{-1}{y-1/y}\times1\\=\frac{-1}{y-1/y}\times\frac yy\\=\frac{-y}{y^2-y/y}=\frac{-y}{y^2-1}\\
=\frac{y}{1-y^2}\]

Answer 25

dy/dx = = ycos(x)/(1-y^2) *

Answer 26

like the original DE

Answer 27

Ahhhhh i see

Answer 28

I think that about does it for that one, will most likely have a few other questions later but thank you for your help

Answer 29

In regard to your question about the absolute value signs in the logarithm,
the different solution sets can be seen attached.

Answer 30

I see thank you

Answer 31

y^2/2 -ln(y) = -sin(x) +1/2

y^2 - ln y^2 = -2sin(x) + 1 (maybe this is the best form for your solution)

Answer 32

alright one other thing
I would like to put into words why the inverse sin of 2 is not a definitive value to use to solve for in this equation, arcsin2=c equation is just an example really need to just explain arcsin2

Answer 33

the sine function take any real input, and returns values between -1 and 1,

Answer 34

the inverse sine function takes values between -1 and +1,
and returns values between -π/2 and +π/2

Answer 35

arcsin(2) asks for the angle thats sine is 2,