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Question Number 181743 by Mastermind last updated on 29/Nov/22

(dy/dx)+xy=xy^2     Solve

$$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{xy}=\mathrm{xy}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Solve} \\ $$

Answered by mr W last updated on 29/Nov/22

(dy/dx)=xy(y−1)  (dy/(y(y−1)))=xdx  ∫(dy/(y(y−1)))=∫xdx  ∫((1/(y−1))−(1/y))dy=∫xdx  ln (y−1)−ln y=((x^2 +C)/2)  ln ((y−1)/y)=((x^2 +C)/2)  1−(1/y)=e^((x^2 +C)/2)   ⇒y=(1/(1−e^((x^2 +C)/2) ))

$$\frac{{dy}}{{dx}}={xy}\left({y}−\mathrm{1}\right) \\ $$$$\frac{{dy}}{{y}\left({y}−\mathrm{1}\right)}={xdx} \\ $$$$\int\frac{{dy}}{{y}\left({y}−\mathrm{1}\right)}=\int{xdx} \\ $$$$\int\left(\frac{\mathrm{1}}{{y}−\mathrm{1}}−\frac{\mathrm{1}}{{y}}\right){dy}=\int{xdx} \\ $$$$\mathrm{ln}\:\left({y}−\mathrm{1}\right)−\mathrm{ln}\:{y}=\frac{{x}^{\mathrm{2}} +{C}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\frac{{y}−\mathrm{1}}{{y}}=\frac{{x}^{\mathrm{2}} +{C}}{\mathrm{2}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{y}}={e}^{\frac{{x}^{\mathrm{2}} +{C}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{1}−{e}^{\frac{{x}^{\mathrm{2}} +{C}}{\mathrm{2}}} } \\ $$

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