I Integral-form change of variable in differential equation

Jaime_mc2
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I have the following differential equation, which is the general Sturm-Liouville problem,
$$
\dfrac{d}{dx} \left[ p(x) \dfrac{d\varphi}{dx} \right] + \left[ \lambda w(x) - q(x) \right] \varphi(x) = 0\ ,
$$
and I want to perform the change of variable
$$
x \rightarrow y = \int_a^x \sqrt{\lambda \dfrac{w(t)}{p(t)}}\, dt\ .
$$

I used the fundamental theorem of calculus to get
$$
\dfrac{dy}{dx} = \sqrt{\lambda\dfrac{w(x)}{p(x)}}\ ,
$$
and get the differentiation operator with respect to the new variable as
$$
\dfrac{d}{dx} = \dfrac{dy}{dx}\dfrac{d}{dy} = \sqrt{\lambda\dfrac{w(x)}{p(x)}} \dfrac{d}{dy}\ .
$$

My question is about how do I get the differential equation to be expressed in terms of ##y##. My first thought was realising that ##\varphi(x)## could just be expressed as ##\varphi(y)## if there exists an inversion such that ##x = x(y)##. Then I just extended this idea to ##w##, ##p## and ##q##, and I arrived to the equation
$$
\dfrac{d}{dy} \left[ \sqrt{w(y)p(y)} \dfrac{d\varphi}{dy} \right] + \left[ \sqrt{w(y)p(y)} - \dfrac{q(y)}{\lambda} \sqrt{\dfrac{p(y)}{w(y)}} \right] \varphi(y) = 0\ .
$$

Is this approach correct?
 
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I observe if you choose w as w=q, the equation becomes easy enough to solve.
 
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