Question

\[(1-x^2)y'-xy=1\]

Answer 1

\[y'-{x \over 1-x^2} y={1 \over (1-x^2)}\]

Answer 2

integration factor : \[ e^{\int {x \over (1-x^2)}dx}\] \[ =e^{{-1 \over 2}ln(1-x^2)} \] \[ = (1-x^2)^{-1/2}\]

Answer 3

\[[y (1-x^2)^{-1/2}]' = (1-x^2)^{-1/2} {1 \over (1-x^2)}\]

Answer 4

\[y(1-x^2)^{-1/2}=\int (1-x^2)^{-3/2}dx\]

Answer 5

\[ y(1-x^2)^{-1/2}={x \over \sqrt{1-x^2} }+c\]

Answer 6

\[ y=x+ c\sqrt{1-x^2 }\]

Answer 7

any mistakes?

Answer 8

\[ y(x) = c/\sqrt{x ^2 -1} - (\ln \sqrt{x ^2 -1} + x)/\sqrt{x ^2 -1} \]

Answer 9

It's very close to being right but I think there's a typo in there somewhere, because if you substitute into the original equation it only works if thats +xy not -xy...

Answer 10

Yes, a couple of mistakes. When you found your integrating factor, you left out a minus sign in the exponent, so it should be \[\sqrt{1-x^2}\] This also changes what will be under the integral in your fourth comment. Once you fix that, you should be okay.

Answer 11

Ah. There it is. Integrating factor should be
\[ e^{\int \frac{-x}{1-x^2} dx} \]

Answer 12

Thank you three,
That is my mistake for sure.
hopefully that is the main(/only) one

Answer 13

is
\[ ln (\sqrt {(x^2-1}))+x = arcsin(x) \]

Answer 14

?

Answer 15

I don't think so.Let me check..........

Answer 16

No, those two things are not the same. Consider x = .5. Then arcsin(x)=pi/6, but the left side clearly does not equal that.

Answer 17

An easier one to check is x = 0

Answer 18

when x=0
they are both 0

Answer 19

When x = 0 LHS = ln (-1) RHS = 0

Answer 20

ops hahah
you re totally right

Answer 21

No Problem.Always happy to help!

Answer 22

i will solve this problem tonight

Answer 23

If you solve the abovementioned equality you will get :
x ~~ -1.79828-3.03269 i...
x ~~ -1.79828+3.03269 i...
x ~~ 1.26215-0.307064 i...
x ~~ 1.26215+0.307064 i...

Answer 24

so they are outta phase?

Answer 25

What do u mean?

Answer 26

the complex bit is negative

Answer 27

Yeah they can be pretty confusing sometimes u have a plus and minus!

Answer 28

\[(1-x^2)y'-xy=1\]\[[y(1-x^2)^{1/2}]'=(1-x^2)^{-1/2}\]\[y(1-x^2)^{1/2}=\int(1-x^2)^{-1/2}dx\]\[y(1-x^2)^{1/2}=arcsin(x)+c\]\[y={arcsin(x) \over \sqrt{1-x^2}} +{c \over \sqrt{1-x^2}}\]

Answer 29

where am i going wrong?

Answer 30

The problem heavily depends on whether you are considering sqrt(1-x^2) or sqrt(x^2-1). Note that the first is only defined for x between -1 and 1 (exclusive, since you eventually divide by it) while the latter is defined for all other values of x (except for 1 and -1, since that would also eventually lead to dividing by zero). Thus, you actually have two very different problems, depending on which region you are considering. You have the correct solution for the interval from -1 to 1. Aron's solution is for the other region, since it involves integrating 1/sqrt(x^2-1), which has a very different solution than the integral of 1/sqrt(1-x^2). Both are correct in their respective regions.

Answer 31

Rewrite the equation:
( dy(x))/( dx)+(x y(x))/(x^2-1) = -1/(x^2-1)
Let mu(x) = exp( integral x/(x^2-1) dx) = sqrt(x^2-1).
Multiply both sides by mu(x):
sqrt(x^2-1) ( dy(x))/( dx)+(x y(x))/sqrt(x^2-1) = -1/sqrt(x^2-1)
Substitute x/sqrt(x^2-1) = ( d)/( dx)(sqrt(x^2-1)):
sqrt(x^2-1) ( dy(x))/( dx)+( d)/( dx)(sqrt(x^2-1)) y(x) = -1/sqrt(x^2-1)
Apply the reverse product rule f ( dg)/( dx)+( df)/( dx) g = ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(sqrt(x^2-1) y(x)) = -1/sqrt(x^2-1)
Integrate both sides with respect to x:
integral ( d)/( dx)(sqrt(x^2-1) y(x)) dx = integral -1/sqrt(x^2-1) dx
Evaluate the integrals:
sqrt(x^2-1) y(x) = -log(x+sqrt(x^2-1))+c_1, where c_1 is an arbitrary constant.
Divide both sides by mu(x) = sqrt(x^2-1):
y(x) = (-log(x+sqrt(x^2-1))+c_1)/sqrt(x^2-1)

Answer 32

@imperialist: great answer, you has explained the issues i was having,
who would have guessed that is two solutions depending on how you handle with negative signs.
i just checked the back of my book and the answer given was the one for the -1 to +1 region, which is what i had.
@Aron West: thank you very much for typing out that second solution for the exterior region.