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Question Number 103412 by abdomsup last updated on 14/Jul/20

solve (1+x^2 )^2 y^(′′) +2x(1+x^2 )y^′ +2=0

$${solve}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} {y}^{''} \:+\mathrm{2}{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+\mathrm{2}=\mathrm{0} \\ $$

Answered by OlafThorendsen last updated on 15/Jul/20

(1+x^2 )^2 y^(′′) +2x(1+x^2 )y^′ +2 = 0 y′ = (u/(1+x^2 )) ⇒ y′′ = (((1+x^2 )u′−2xu)/((1+x^2 )^2 )) (1+x^2 )u′−2xu+2xu+2 = 0 u′ = −(2/(1+x^2 )) ⇒u = −2Arctanx+C_1 y′ = −2((Arctanx)/(1+x^2 ))+(C_1 /(1+x^2 )) ⇒ y = −Arctan^2 x+C_1 Arctanx+C_2

$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} {y}^{''} \:+\mathrm{2}{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+\mathrm{2}\:=\:\mathrm{0} \\ $$$${y}'\:=\:\frac{{u}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{y}''\:=\:\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){u}'−\mathrm{2}{xu}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){u}'−\mathrm{2}{xu}+\mathrm{2}{xu}+\mathrm{2}\:=\:\mathrm{0} \\ $$$${u}'\:=\:−\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}\:=\:−\mathrm{2Arctan}{x}+\mathrm{C}_{\mathrm{1}} \\ $$$${y}'\:=\:−\mathrm{2}\frac{\mathrm{Arctan}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{C}_{\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{y}\:=\:−\mathrm{Arctan}^{\mathrm{2}} {x}+\mathrm{C}_{\mathrm{1}} \mathrm{Arctan}{x}+\mathrm{C}_{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 15/Jul/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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