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Question Number 128106 by bemath last updated on 04/Jan/21

yy′ + b(x−a)= ((ay^2 )/(1+x^2 ))

$$\:\mathrm{yy}'\:+\:\mathrm{b}\left(\mathrm{x}−\mathrm{a}\right)=\:\frac{\mathrm{ay}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\: \\ $$

Commented by mr W last updated on 05/Jan/21

once i asked this question due to Q127997, but then i could solve it myself and therefore have deleted the post. my post existed only for a short time. thank you that you have noticed my post nevertheless and restored the question here! i enjoyed that i could solve Q127997 completely. please check out the solution there, comments are welcome!

$${once}\:{i}\:{asked}\:{this}\:{question}\:{due}\:{to} \\ $$$${Q}\mathrm{127997},\:{but}\:{then}\:{i}\:{could}\:{solve}\:{it}\:{myself} \\ $$$${and}\:{therefore}\:{have}\:{deleted}\:{the}\:{post}. \\ $$$${my}\:{post}\:{existed}\:{only}\:{for}\:{a}\:{short}\:{time}. \\ $$$${thank}\:{you}\:{that}\:{you}\:{have}\:{noticed}\:{my}\: \\ $$$${post}\:{nevertheless}\:{and}\:{restored}\:{the} \\ $$$${question}\:{here}! \\ $$$${i}\:{enjoyed}\:{that}\:{i}\:{could}\:{solve} \\ $$$${Q}\mathrm{127997}\:{completely}.\:{please}\:{check}\:{out} \\ $$$${the}\:{solution}\:{there},\:{comments}\:{are} \\ $$$${welcome}! \\ $$

Answered by liberty last updated on 04/Jan/21

yy′+b(x−a)=((ay^2 )/(1+x^2 )) yy′−((ay^2 )/(1+x^2 )) = b(a−x) let y^2 = v ⇒2yy′ = v′ ; yy′ = ((v′)/2) ⇒ ((v′)/2)−(a/(1+x^2 )).v = b(a−x) ; v′−((2a)/(1+x^2 )).v=2b(a−x) put IF μ = e^(−∫ ((2a)/(1+x^2 )) dx) = e^(−2a .arctan x) we get v = ((∫2b(a−x)e^(−2a.arctan x) dx+C)/e^(−2.arctan x) ) ∴ y^2 = C.e^(2a.arctan x) + 2b∫ (a−x).e^(−2a.arctan x) dx

$$\:\mathrm{yy}'+\mathrm{b}\left(\mathrm{x}−\mathrm{a}\right)=\frac{\mathrm{ay}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\mathrm{yy}'−\frac{\mathrm{ay}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{b}\left(\mathrm{a}−\mathrm{x}\right) \\ $$$$\:\mathrm{let}\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{v}\:\Rightarrow\mathrm{2yy}'\:=\:\mathrm{v}'\:;\:\mathrm{yy}'\:=\:\frac{\mathrm{v}'}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{v}'}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\mathrm{v}\:=\:\mathrm{b}\left(\mathrm{a}−\mathrm{x}\right)\:;\:\mathrm{v}'−\frac{\mathrm{2a}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\mathrm{v}=\mathrm{2b}\left(\mathrm{a}−\mathrm{x}\right) \\ $$$$\mathrm{put}\:\mathrm{IF}\:\mu\:=\:\mathrm{e}^{−\int\:\frac{\mathrm{2a}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}} =\:\mathrm{e}^{−\mathrm{2a}\:.\mathrm{arctan}\:\mathrm{x}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{v}\:=\:\frac{\int\mathrm{2b}\left(\mathrm{a}−\mathrm{x}\right)\mathrm{e}^{−\mathrm{2a}.\mathrm{arctan}\:\mathrm{x}} \:\mathrm{dx}+\mathrm{C}}{\mathrm{e}^{−\mathrm{2}.\mathrm{arctan}\:\mathrm{x}} } \\ $$$$\:\therefore\:\mathrm{y}^{\mathrm{2}} =\:\mathrm{C}.\mathrm{e}^{\mathrm{2a}.\mathrm{arctan}\:\mathrm{x}} \:+\:\mathrm{2b}\int\:\left(\mathrm{a}−\mathrm{x}\right).\mathrm{e}^{−\mathrm{2a}.\mathrm{arctan}\:\mathrm{x}} \:\mathrm{dx}\: \\ $$

Commented by mr W last updated on 05/Jan/21

for complete solution see Q127997.

$${for}\:{complete}\:{solution}\:{see}\:{Q}\mathrm{127997}. \\ $$

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