Changing to polar coordinates bring the differential equation to Φ(ϕ,ρ,ρ′(ϕ))=0 form.I have a differential equation y′=dx+yx−y and the problem says to change to polar coordinate system by assuming x=ρcos⁡ϕ,y=ρsin⁡ϕ and ρ=ρ(ϕ) and bring the equation to Φ(ϕ,ρ,ρ′(ϕ))=0 form. The answer is ρ′(ϕ)=ρ(ϕ). This problem is under "Implicit Functions" in the book, so I suppose I should use the Implicit-function theorem. But I don't see what I can get from that. I did substitute and tried to simplify, but in vain. The only way I see to solve this after substitution is by applying trigonometry, but every time either ρ(ϕ) orρ′(ϕ) are multiplied by some other trigonometric function (that can't be simplified to a constant) and it seems that just trigonometry is not enough. Could you explain how such problems are solved? I couldn't find examples on the internet either.

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Changing to polar coordinates bring the differential equation to Φ(ϕ,ρ,ρ′(ϕ))=0 form.I have a differential equation y′=dx+yx−y and the problem says to change to polar coordinate system by assuming x=ρcos⁡ϕ,y=ρsin⁡ϕ and ρ=ρ(ϕ) and bring the equation to Φ(ϕ,ρ,ρ′(ϕ))=0 form. The answer is ρ′(ϕ)=ρ(ϕ). This problem is under &quot;Implicit Functions&quot; in the book, so I suppose I should use the Implicit-function theorem. But I don&#039;t see what I can get from that. I did substitute and tried to simplify, but in vain. The only way I see to solve this after substitution is by applying trigonometry, but every time either ρ(ϕ) orρ′(ϕ) are multiplied by some other trigonometric function (that can&#039;t be simplified to a constant) and it seems that just trigonometry is not enough. Could you explain how such problems are solved? I couldn&#039;t find examples on the internet either.

Meadow Franco · Accepted Answer

Step 1Multiply out the numerator and re-arrange to getxy′−y=x+yy′or more suggestive x,dy−y,dx=x,dx+y,dyStep 2Under polar coordinates this reads also asρ2,dϕ=ρ,dρOne can reason that in most situations this can be canceled todρdϕ=ρwith an implied change in the independent variable.

massifyc9 · Accepted Answer

Step 1y′=dx+yx−yddydx=dx+yx−yBe carefull and apply the chain rule:ddydtddtdx=dx+yx−yddydt=ddxdtdx+yx−yStep 2Substitute now y=r(t)sin⁡t:ddr(t)sin⁡tdt=dcos⁡t+sin⁡tcos⁡t−sin⁡tddr(t)cos⁡(t)dtr′sin⁡t+rcos⁡t=dcos⁡t+sin⁡tcos⁡t−sin⁡t(r′cos⁡t−rsin⁡t)(r′sin⁡t+rcos⁡t)(cos⁡t−sin⁡t)=(cos⁡t+sin⁡t)(r′cos⁡t−rsin⁡t)This is simple trigonometry. It should simplify nicely and give you Lutz Lehmann&#039;s nice answer.r′=r

Changing to polar coordinates bring the differential equation

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