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Question Number 126229 by MathSh last updated on 18/Dec/20
(1+y)x′=x
$$\left(\mathrm{1}+{y}\right){x}'={x} \\ $$
Answered by Dwaipayan Shikari last updated on 18/Dec/20
(1+y)(dx/dy)=x ⇒∫(dy/(1+y))=∫(dx/x) ⇒1+y=Cx
$$\left(\mathrm{1}+{y}\right)\frac{{dx}}{{dy}}={x}\:\Rightarrow\int\frac{{dy}}{\mathrm{1}+{y}}=\int\frac{{dx}}{{x}}\:\Rightarrow\mathrm{1}+{y}={Cx} \\ $$
Answered by ebi last updated on 18/Dec/20
(y+1)x′=x  ((1/x))x′=(1/(y+1))  ∫((1/x))dx=∫(1/(y+1))dy  ln∣x∣=ln∣y+1∣+c  ln∣x∣=ln∣y+1∣+ln∣C∣  ln∣x∣=ln∣C(y+1)∣  x=C(y+1)
$$\left({y}+\mathrm{1}\right){x}'={x} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right){x}'=\frac{\mathrm{1}}{{y}+\mathrm{1}} \\ $$$$\int\left(\frac{\mathrm{1}}{{x}}\right){dx}=\int\frac{\mathrm{1}}{{y}+\mathrm{1}}{dy} \\ $$$${ln}\mid{x}\mid={ln}\mid{y}+\mathrm{1}\mid+{c} \\ $$$${ln}\mid{x}\mid={ln}\mid{y}+\mathrm{1}\mid+{ln}\mid{C}\mid \\ $$$${ln}\mid{x}\mid={ln}\mid{C}\left({y}+\mathrm{1}\right)\mid \\ $$$${x}={C}\left({y}+\mathrm{1}\right) \\ $$

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