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Question Number 161058 by HongKing last updated on 11/Dec/21

Solve the differential equation: x(y-1)dx + (x+1)dy = 0

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\mathrm{x}\left(\mathrm{y}-\mathrm{1}\right)\mathrm{dx}\:+\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{dy}\:=\:\mathrm{0} \\ $$$$ \\ $$

Answered by mr W last updated on 11/Dec/21

(dy/dx)+(x/(x+1))(y−1)=0 ((d(y−1))/dx)+(x/(x+1))(y−1)=0 ((d(y−1))/(y−1))=((1/(x+1))−1)dx ∫((d(y−1))/(y−1))=∫((1/(x+1))−1)dx ln (y−1)=ln (x+1)−(x+C_1 ) ln ((y−1)/(x+1))=−(x+C_1 ) ((y−1)/(x+1))=(1/(Ce^x )) ⇒y=((x+1)/(Ce^x ))+1

$$\frac{{dy}}{{dx}}+\frac{{x}}{{x}+\mathrm{1}}\left({y}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{d}\left({y}−\mathrm{1}\right)}{{dx}}+\frac{{x}}{{x}+\mathrm{1}}\left({y}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{d}\left({y}−\mathrm{1}\right)}{{y}−\mathrm{1}}=\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{1}\right){dx} \\ $$$$\int\frac{{d}\left({y}−\mathrm{1}\right)}{{y}−\mathrm{1}}=\int\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{1}\right){dx} \\ $$$$\mathrm{ln}\:\left({y}−\mathrm{1}\right)=\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\left({x}+{C}_{\mathrm{1}} \right) \\ $$$$\mathrm{ln}\:\frac{{y}−\mathrm{1}}{{x}+\mathrm{1}}=−\left({x}+{C}_{\mathrm{1}} \right) \\ $$$$\frac{{y}−\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{1}}{{Ce}^{{x}} } \\ $$$$\Rightarrow{y}=\frac{{x}+\mathrm{1}}{{Ce}^{{x}} }+\mathrm{1} \\ $$

Commented by HongKing last updated on 11/Dec/21

very nice dear Sir thank you so much

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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