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Question Number 103168 by mohammad17 last updated on 13/Jul/20

The particular solution of differential equation   of (dy/dx)+(y/x)=k is y=(1/x)+2x thus ,whats the value of k ?

$${The}\:{particular}\:{solution}\:{of}\:{differential}\:{equation}\: \\ $$$${of}\:\frac{{dy}}{{dx}}+\frac{{y}}{{x}}={k}\:{is}\:{y}=\frac{\mathrm{1}}{{x}}+\mathrm{2}{x}\:{thus}\:,{whats}\:{the}\:{value}\:{of}\:{k}\:? \\ $$

Answered by bemath last updated on 13/Jul/20

IF u(x)= e^(∫ (dx/x))  = e^(ln(x))  = x  x.(dy/dx) + y = kx   (d/dx) (xy) = kx   xy = (1/2)kx^2  + C   y = ((kx)/2) + (C/x) , then ((kx)/2) = 2x  ∴ k = 4

$${IF}\:{u}\left({x}\right)=\:{e}^{\int\:\frac{{dx}}{{x}}} \:=\:{e}^{\mathrm{ln}\left({x}\right)} \:=\:{x} \\ $$$${x}.\frac{{dy}}{{dx}}\:+\:{y}\:=\:{kx}\: \\ $$$$\frac{{d}}{{dx}}\:\left({xy}\right)\:=\:{kx}\: \\ $$$${xy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \:+\:{C}\: \\ $$$${y}\:=\:\frac{{kx}}{\mathrm{2}}\:+\:\frac{{C}}{{x}}\:,\:{then}\:\frac{{kx}}{\mathrm{2}}\:=\:\mathrm{2}{x} \\ $$$$\therefore\:{k}\:=\:\mathrm{4}\: \\ $$

Commented by mohammad17 last updated on 13/Jul/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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