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

Question

yaum3xg1 · 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;,, and&amp;nbsp;f&#039;=(F&#039;∘y)·y&#039;,, 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;.

Riya Erickson · Accepted Answer

Step 1It is correct thatddxp·dfdx=ddyp·dFdy·dydx·dydx.Here, you must use the product rule:ddyp·dFdy·dydx·dydx=ddyp·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 \(\displaystyle{d}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({p}{\left({x}\right)}{d}{\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}}\right)}+\cdots={0}\) and I want

Answered question

Answer & Explanation

New Questions in Calculus 2