Question

Trying to solve another First order linear ODE using an integrating factor; workings posted below shortly.

Answer 1

\[2(1-x)y'-y=4x\sqrt{1-x}, \ \ \ y(0)=-1.\]

Answer 2

\[\frac{dy}{dx}-\frac{y}{2(1-x)}=\frac{4x \sqrt{1-x}}{2(1-x)}\]

Answer 3

\[\mu(x) \frac{dy}{dx}-\mu(x)\frac{y}{2(1-x)}=\mu(x) \frac{4x \sqrt{1-x}}{2(1-x)}\]

Answer 4

\[\mu(x)=e^{\int\limits_{}^{}P(x)dx}=e^{- \int\limits_{}^{}\frac{1}{2(1-x)}dx}\]

Answer 5

\[\frac{1}{2}\int\limits_{}^{}\frac{-1}{(1-x)}dx=\frac{1}{2}\ln(1-x) +C\]
Getting rid of the constant of integration in this context,

Answer 6

\[\mu(x)=e^{\frac{1}{2}\ln(1-x)}=\sqrt{1-x}\]

Answer 7

\[f(x)\mu(x)=\frac{2x}{\sqrt{1-x}}\sqrt{1-x}=2x\]

Answer 8

\[y=\frac{\int\limits_{}^{}f(x)\mu(x)}{\mu(x)}dx=\frac{\int\limits_{}^{}2x \ dx}{\sqrt{1-x}}=\frac{x^2+C}{\sqrt{1-x}}\]

Answer 9

Sticking in the conditions applied to the IVP, \[y(0)=-1=\frac{(0)^2+C}{\sqrt{1-(0)}}=C; \ \ \ C = -1.\]

Answer 10

Answer is supposed to be:

Answer 11

Hint :
\[x^2 - 1 = -(1-x^2) = -(1+x)(1-x)\]

Answer 12

What they did was:\[
\frac{1-x}{\sqrt{1-x}} = \sqrt{1-x}
\]

Answer 13

Yeah, I got it now. Thanks, nonetheless.