so ive put lnk on the side with x but with this i get k=-1 which is wrong
#what side does lnk go
dusky blade
so you've found the general solution?
but you cannot find the constant term to get the particular one?
prime anvil
yeh ik the solution but instead of putting lnk on the right hand side they have put it on the left
which gives k=1
dusky blade
when intergrated it is -ln(1-y^2)=ln(x+1)+lnk
I don't think that is the form they are looking for
prime anvil
-(1+y^2)=k(x+1)
dusky blade
probably you just wolfram-alpha-ed this XD
prime anvil
when x is 0 and y is 0 -1=k
dusky blade
oh
prime anvil
then when you sub k back in to get it into y=g(x) u get y^2=-x
which is not correct
maybe there is an error when dividing by ln
idk if u can divide ln by -ln to get -
prime anvil
when intergrated it is -ln(1-y^2)=ln(x+1)+lnk
i combined this
so its lnk(x+1) on the right side
dusky blade
because ln(1-y^2) = 0 when y = 0 and so is ln(x+1) when x = 0
prime anvil
then divide whole thing by ln
dusky blade
you can't divide by ln it's a function
I am asking you for the full expression please
prime anvil
wdym full expression
prime anvil
when intergrated it is -ln(1-y^2)=ln(x+1)+lnk
this is it
prime anvil
ln
dusky blade
it's not correct
you cannot divide by ln any more than you can divide by "+"
it's an operation not a number
ln(x) is the natural logarithm
aka log_e(x)
(log with base e)
does this speak to you?
prime anvil
this is the answer
im confused about which side the lnk goes to the left or the right as it always changes when i do these questions
dusky blade
does this speak to you?
?
prime anvil
er ig
prime anvil
im confused about which side the lnk goes to the left or the right as it always ...
what about this tho
dusky blade
can you divide by "ln" ?
prime anvil
no
dusky blade
great
dusky blade
and this step is not in the solution above? There is no division by an expression with "ln"
prime anvil
yeh it was removing ln
dusky blade
there is no such thing as "removing ln"
dusky blade
you mean lines 3, 4?
dusky blade
it "dissappears" because ln(a) - ln(b) = ln(a/b)
(b=/=0)
and as far as I am concerned it does not matter where the k is
prime anvil
yeh i thought so too but if i put it on the right i get k is=-1
dusky blade
idk what this even means but you probably just mess up with some arithmetic
lone lotus
im confused about which side the lnk goes to the left or the right as it always ...
It doesn't matter.
prime anvil
when intergrated it is -ln(1-y^2)=ln(x+1)+lnk
.
prime anvil
-(1+y^2)=k(x+1)
.
prime anvil
when x is 0 and y is 0 -1=k
.
prime anvil
idk what this even means but you probably just mess up with some arithmetic
ig but i cant spot the mistake
lone lotus
Again, it doesn't matter what you do with k. The final function after you applied the initial condition will be the same, anyway.
dusky blade
ig but i cant spot the mistake
show us the working then
lone lotus
Let me try.
dusky blade
nah, @lone lotus ofc you won't make the mistake
he needs to do it for us to spot his
prime anvil
look up
prime anvil
thats not my working
dusky blade
so
prime anvil
when intergrated it is -ln(1-y^2)=ln(x+1)+lnk
this is
prime anvil
-(1+y^2)=k(x+1)
..
prime anvil
when x is 0 and y is 0 -1=k
..
dusky blade
well
that's the mistake
first line
you did not multiply the term by (-1) when "changing sides"
really you are subtracting ln(k) from both sides
prime anvil
?
dusky blade
but you subtracted it from LHS and added on the RHS
lone lotus
2y(1 + x)dy = (1 - y^2)dx
2ydy/(y^2 - 1) = -dx/(1 + x)
We integrate.
ln(|1 - y^2|) = ln(|k|) - ln(|1 + x|)
1 - y^2 = k/(1 + x)
y^2 = 1 - k/(1 + x)
We have y(0) = 0. So:
0 = 1 - k
k = 1
Thus, the solution is:
y^2 = 1 - 1/(1 + x)
y^2 = x/(1 + x)
You can split it into y = ±√(x/(1 + x)), or write it like x = y^2/(1 - y^2), doesn't matter.
prime anvil
cant u just leave the minus there
dusky blade
2y(1 + x)dy = (1 - y^2)dx
2ydy/(y^2 - 1) = -dx/(1 + x)
We integrate.
ln(\|1 - y^...
impressively concise and fast working out
but we kinda had it already and OP was searching for his error
as opposed to a solution
prime anvil
but you subtracted it from LHS and **added** on the RHS
subtracted what
dusky blade
ln(k)
dusky blade
look at the third line here
prime anvil
yeh but this ln k can be on either side
prime anvil
2y(1 + x)dy = (1 - y^2)dx
2ydy/(y^2 - 1) = -dx/(1 + x)
We integrate.
ln(\|1 - y^...
so the reason mine doesnt work is because there is a minus before the ln?
lone lotus
so the reason mine doesnt work is because there is a minus before the ln?
Where is your solution?
dusky blade
the correct line that is closest to your line is
ln(|k|)-ln(|1-y^2|)=ln(|1+x|)
now you make the error of "bringing" the ln(k) to the other side without flipping the sign
prime anvil
ye ig that but the lnk can go on the other side without flipping the sign
@lone lotus
lone lotus
-ln(1-y^2)=ln(1+x) +lnk
-ln(1-y^2)=lnk+kx
-ln(1 - y^2) = ln(k + kx) is correct, but you potentiated wrong. The result should be:
1/(1 - y^2) = k + kx
Remember: e^(-ln(x)) = 1/x, not -x.
prime anvil
-ln(1 - y^2) = ln(k + kx) is correct, but you potentiated wrong. The result shou...
👍 yeh makes sense
bring the minus one to the power
thanks
dusky blade
bring the minus one to the power
no?
dusky blade
it's to take to a negative power
constant of intergration, right
ok
well, I think you are good now. feel free to come back
prime anvil
thanks
dusky blade
it was @lone lotus that spotted what you needed after all but welcome :)