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Question-139445




Question Number 139445 by aliibrahim1 last updated on 27/Apr/21
Answered by mr W last updated on 27/Apr/21
u=xy  (du/dx)=x(dy/dx)+y  (du/dx)+3u^2 =0  −(du/u^2 )=3dx  (1/u)=3x+C  (1/(xy))=3x+C  (1/(2×(1/2)))=3×2+C ⇒C=−5  (1/(xy))=3x−5  y=(1/(x(3x−5)))  y(1)=(1/(1×(3×1−5)))=−(1/2)
$${u}={xy} \\ $$$$\frac{{du}}{{dx}}={x}\frac{{dy}}{{dx}}+{y} \\ $$$$\frac{{du}}{{dx}}+\mathrm{3}{u}^{\mathrm{2}} =\mathrm{0} \\ $$$$−\frac{{du}}{{u}^{\mathrm{2}} }=\mathrm{3}{dx} \\ $$$$\frac{\mathrm{1}}{{u}}=\mathrm{3}{x}+{C} \\ $$$$\frac{\mathrm{1}}{{xy}}=\mathrm{3}{x}+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3}×\mathrm{2}+{C}\:\Rightarrow{C}=−\mathrm{5} \\ $$$$\frac{\mathrm{1}}{{xy}}=\mathrm{3}{x}−\mathrm{5} \\ $$$${y}=\frac{\mathrm{1}}{{x}\left(\mathrm{3}{x}−\mathrm{5}\right)} \\ $$$${y}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{1}×\left(\mathrm{3}×\mathrm{1}−\mathrm{5}\right)}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by aliibrahim1 last updated on 27/Apr/21
thx bro
$${thx}\:{bro} \\ $$

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