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Question Number 102515 by bemath last updated on 09/Jul/20

((x/y))y′= ((2y^2 +1)/(x+1))

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

Answered by bobhans last updated on 09/Jul/20

(dy/(y(2y^2 +1))) = (dx/(x(x+1))) (1) (1/(y(2y^2 +1))) = (A/y) + ((By+C)/(2y^2 +1)) 1 = A(2y^2 +1)+(By+C)y ⇒ { ((A=1)),((1=3.1−(−B+C))),((1=3.1+B+C)) :} −2 = B−C ∧−2=B+C ⇒B=−2 ; C=0 ∫(dy/y)−∫((2y dy )/(2y^2 +1)) = ∫ (dx/x) −∫ (dx/(x+1)) ln(y)−(1/2)ln(2y^2 +1) = ln∣C((x/(x+1)))∣ ln∣(y/(√(2y^2 +1)))∣ = ln ∣((Cx)/(x+1)) ∣ (y/(√(2y^2 +1))) = ((Cx)/(x+1)) ⊕

$$\frac{{dy}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{A}}{{y}}\:+\:\frac{{By}+{C}}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{1}\:=\:{A}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)+\left({By}+{C}\right){y}\: \\ $$$$\Rightarrow\begin{cases}{{A}=\mathrm{1}}\\{\mathrm{1}=\mathrm{3}.\mathrm{1}−\left(−{B}+{C}\right)}\\{\mathrm{1}=\mathrm{3}.\mathrm{1}+{B}+{C}}\end{cases} \\ $$$$−\mathrm{2}\:=\:{B}−{C}\:\wedge−\mathrm{2}={B}+{C} \\ $$$$\Rightarrow{B}=−\mathrm{2}\:;\:{C}=\mathrm{0} \\ $$$$\int\frac{{dy}}{{y}}−\int\frac{\mathrm{2}{y}\:{dy}\:}{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:\int\:\frac{{dx}}{{x}}\:−\int\:\frac{{dx}}{{x}+\mathrm{1}} \\ $$$$\mathrm{ln}\left({y}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)\:=\:\mathrm{ln}\mid{C}\left(\frac{{x}}{{x}+\mathrm{1}}\right)\mid\: \\ $$$$\mathrm{ln}\mid\frac{{y}}{\sqrt{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}}\mid\:=\:\mathrm{ln}\:\mid\frac{{Cx}}{{x}+\mathrm{1}}\:\mid\: \\ $$$$\frac{{y}}{\sqrt{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}}\:=\:\frac{{Cx}}{{x}+\mathrm{1}}\:\oplus\: \\ $$

Answered by petrochengula last updated on 09/Jul/20

(dy/(y(2y^2 +1)))=(dx/(x(x+1))) integrate both sides ∫(dy/(y^3 (2+(1/y^2 ))))=∫(dx/(x^2 (1+(1/x)))) −(1/2)ln(((2y^2 +1)/y^2 ))=ln∣((Cx)/(1+x))∣+ ln(√(y^2 /(2y^2 +1)))=ln∣((Cx)/(1+x))∣ (y/(√(2y^2 +1)))=((Cx)/(1+x))

$$\frac{{dy}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)} \\ $$$${integrate}\:{both}\:{sides} \\ $$$$\int\frac{{dy}}{{y}^{\mathrm{3}} \left(\mathrm{2}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)}=\int\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{{y}^{\mathrm{2}} }\right)={ln}\mid\frac{{Cx}}{\mathrm{1}+{x}}\mid+ \\ $$$${ln}\sqrt{\frac{{y}^{\mathrm{2}} }{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}}={ln}\mid\frac{{Cx}}{\mathrm{1}+{x}}\mid \\ $$$$\frac{{y}}{\sqrt{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}}=\frac{{Cx}}{\mathrm{1}+{x}} \\ $$

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