I have the differential equationddx(p(x)dfdx)+⋯=0and I want to perform a generic change of variable from x to y=y(x).

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Kendrick Fritz · Accepted Answer

Step 1It may be illuminating to work using functional notation rather than Leibniz notation. The expression you are concerned with rewriting is(p·f&#039;)&#039;=p&#039;·f&#039;+p·f&#039;&#039;.Let&amp;nbsp;p=P∘y&amp;nbsp;and&amp;nbsp;f=F∘y.. Thus&amp;nbsp;p&#039;=(P&#039;∘y)·y&#039;,&amp;nbsp;and&amp;nbsp;f&#039;=(F&#039;∘y)·y&#039;,&amp;nbsp;and&amp;nbsp;f&#039;&#039;=(F&#039;∘y)&#039;·y&#039;+(F&#039;∘y)·y&#039;&#039;=(F&#039;&#039;∘y)·(y&#039;)2+(F&#039;∘y)·y&#039;&#039;.As such,(p·f&#039;)&#039;=(P&#039;∘y)·y&#039;·(F&#039;∘y)·y&#039;+(P∘y)·(F&#039;&#039;∘y)·(y&#039;)2+(P∘y)·(F&#039;∘y)·y&#039;&#039;.

obettyQuokeperg6 · Accepted Answer

Step 1It is correct thatddx[p·dfdx]=ddy[p·dFdy·dydx]·dydx.Here, you must use the product rule:ddy[p·dFdy·dydx]·dydx=ddy[p·dFdy](dydx)2+p·dFdy·dydx·ddy(dydx)The secret here is thatddy(dydx)=ddy(1dxdy)=-d2xdy2(dydx)2.The problem here is that, in actuality, you really cannot do such arbitrary changes of variable without accounting for y&#039; in the change of variables.

I have the differential equation \frac{d}{dx} ( p(x) \frac{df}{dx} ) +

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