Menu Close

Question-98808




Question Number 98808 by bemath last updated on 16/Jun/20
Answered by john santu last updated on 16/Jun/20
y = wx ⇒ (dy/dx) = w + x (dw/dx)  ⇔w + x (dw/dx) = ((wx^2 −w^2 x^2 )/(x^2 +wx^2 ))  x (dw/dx)  = ((w−w^2 )/(1+w)) − w = ((−2w^2 )/(1+w))  (((1+w) dw)/w^2 ) = −2 (dx/x)  ∫ (w^(−2) +w^(−1) )dw = −2∫ (dx/x)  −(1/w) + ln ∣w∣ = −2 ln ∣x∣ + c  ln ∣ (w/x^2 )∣ = (1/w) + c   (y/(x^3  )) = e^((1/x) + c) ⇔ y = x^3  (Ce^(1/x) )
$${y}\:=\:{wx}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{w}\:+\:{x}\:\frac{{dw}}{{dx}} \\ $$$$\Leftrightarrow{w}\:+\:{x}\:\frac{{dw}}{{dx}}\:=\:\frac{{wx}^{\mathrm{2}} −{w}^{\mathrm{2}} {x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{wx}^{\mathrm{2}} } \\ $$$${x}\:\frac{{dw}}{{dx}}\:\:=\:\frac{{w}−{w}^{\mathrm{2}} }{\mathrm{1}+{w}}\:−\:{w}\:=\:\frac{−\mathrm{2}{w}^{\mathrm{2}} }{\mathrm{1}+{w}} \\ $$$$\frac{\left(\mathrm{1}+{w}\right)\:{dw}}{{w}^{\mathrm{2}} }\:=\:−\mathrm{2}\:\frac{{dx}}{{x}} \\ $$$$\int\:\left({w}^{−\mathrm{2}} +{w}^{−\mathrm{1}} \right){dw}\:=\:−\mathrm{2}\int\:\frac{{dx}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{w}}\:+\:\mathrm{ln}\:\mid{w}\mid\:=\:−\mathrm{2}\:\mathrm{ln}\:\mid{x}\mid\:+\:\mathrm{c} \\ $$$$\mathrm{ln}\:\mid\:\frac{{w}}{{x}^{\mathrm{2}} }\mid\:=\:\frac{\mathrm{1}}{{w}}\:+\:{c}\: \\ $$$$\frac{{y}}{{x}^{\mathrm{3}} \:}\:=\:{e}^{\frac{\mathrm{1}}{{x}}\:+\:{c}} \Leftrightarrow\:\mathrm{y}\:=\:{x}^{\mathrm{3}} \:\left({Ce}^{\frac{\mathrm{1}}{{x}}} \right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *