Question

I'm having trouble solving this differential equation

(1 - x^2) y' - xy = x
I'm not sure about the integrating factor

is it (1 - x^2) ^(1/2) or 1/ ((1 - x^2)^(1/2)
(or maybe its neither!). Please help.

Answer 1

if in the form
y' + P(x) y = Q(x)
the integrating factor is e^F
where F= integral P(x) dx

Answer 2

\[(1-x^2)\frac{dy}{dx} -xy = x\]

Answer 3

@ph1 - yes - I remembered that I'm (to my shame) confuse by the algebra

Answer 4

integral -x dx/(1-x^2) = 1/2 * ln(1-x^2) (toss in abs signs to be sure it's positive)

Answer 5

exp (ln(1-x^2)^(1/2))= sqrt(1-x^2)

Answer 6

integral du/u = ln(u)

in this case u = 1-x^2 du = -2x dx

Answer 7

\[ e^{ab} = \left(e^a\right)^b \]
so
\[ e^{\ln(1-x^2)\cdot \frac{1}{2}}=\left(e^{\ln(1-x^2)}\right)^\frac{1}{2} \]
\[= (1-x^2)^\frac{1}{2}= \sqrt{1-x^2} \]

Answer 8

ok - i see it now . I was confused with the sign when integrating
thanx

Answer 9

i need more practice in FODEs .......

Answer 10

are you saying you can't finish ?

Answer 11

no i think i've got it now

I get the solution to be

y = A(1 - x^2)^(-1/2) - 1

Answer 12

that looks good

Answer 13

I got it down to