How to write a non-homogeneous equation in self-adjoint form

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How can I write a non-homogeneous equation in self-dajoint form?

such as, for equation with $-1\le x \le1$

$$(1-x^2)u''-xu'+2u=x^4+x$$

What is its self-dajoint form?

Also, for a homogeneous equation e.g

$$p_0(x)u''+p_1(x)u'+p_2(x)u=0 \quad where \quad x\in I $$

How can I find the $p(x),\rho(x)and q(x)$ in the self-adjoint form?

$$\frac{d}{dx}[p(x)\frac{du}{dx}]+[\lambda \rho(x)-q(x)]u=0$$

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Suppose you have

$$ y'' + f y' + g y = h $$

By means of an integrating factor, $ \mu =\exp (\int f)$, we obtain

$$ \mu \left ( y'' + fy' + gy \right) = \mu h \iff ( \mu y')' + g\mu y = \mu h$$

Thus, in your example, we have

$$ u'' + \frac{x}{x^2-1} u' +\frac{2}{ 1-x^2} u =x\frac{x^3+1}{1-x^2} $$

So we want to choose an integrating factor like

$$ \mu = \exp \left ( \int \frac{x \ dx}{x^2-1} \right ) = \exp \left( \ln\left(\sqrt{ x^2 -1}\right)\right) = \sqrt{ x^2 -1} $$

Thus the self-adjoint form would be

$$ -( \sqrt{x^2-1} u' )' + \frac{2 u}{\sqrt{x^2-1}}= x \frac{x^3 +1}{\sqrt{x^2-1}} $$