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

$${\mathrm{yy}}^{\prime }+\mathrm{b}(\mathrm{x}-\mathrm{a})=\frac{{\mathrm{ay}}^2}{1+{\mathrm{x}}^2}$$

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

$$onceiaskedthisquestiondueto$$$$Q127997,butthenicouldsolveitmyself$$$$andthereforehavedeletedthepost.$$$$mypostexistedonlyforashorttime.$$$$thankyouthatyouhavenoticedmy$$$$postneverthelessandrestoredthe$$$$questionhere!$$$$ienjoyedthaticouldsolve$$$$Q127997completely.pleasecheckout$$$$thesolutionthere,commentsare$$$$welcome!$$

Answered by liberty last updated on 04/Jan/21

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

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

$$forcompletesolutionseeQ127997.$$

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