Question

hello help me solve this equation by the linear method.
(1-x ²) y '+ xy = x

Answer 1

Does it have to be the linear method?

Answer 2

first ODE?

Answer 3

first do this
((1-x ²) dy / dx / (1-x ²)) + (x / (1-x ²)) = (x / (1-x ²))
clears dy / dx

dy / dx + ((x, y) / (1-x ²)) = (x / (1-x ²))
x ² is removed?

Answer 4

\[(1-x^2)y'+xy=x\\
y'+\frac{x}{1-x^2}y=\frac{x}{1-x^2}\]
Integrating factor:
\[\mu(x)=\exp\left(\int\frac{x}{1-x^2}~dx\right)\\
\mu(x)=\exp\left(\ln\left|\frac{1}{\sqrt{1-x^2}}\right|\right)\\
\mu(x)=\frac{1}{\sqrt{1-x^2}}\]
\[\frac{1}{\sqrt{1-x^2}}y'+\frac{x}{\left(1-x^2\right)^{3/2}}y=\frac{x}{\left(1-x^2\right)^{3/2}}\\
\frac{d}{dx}\left[\frac{x}{(1-x^2)^{3/2}}y\right]=\frac{x}{(1-x^2)^{3/2}}\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\vdots \]
Note that the equation is also separable, as I'm sure @blurbendy was thinking.
\[(1-x^2)y'+xy=x~\iff~\frac{1}{1-y}~dy=\frac{x}{1-x^2}~dx\]