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Question Number 51840 by 33 last updated on 31/Dec/18

solve  (dy/(dx )) + ((2y)/(3x )) = (x/(√y))

$${solve} \\ $$$$\frac{{dy}}{{dx}\:}\:+\:\frac{\mathrm{2}{y}}{\mathrm{3}{x}\:}\:=\:\frac{{x}}{\sqrt{{y}}} \\ $$

Answered by ajfour last updated on 31/Dec/18

3x(√y)dy+2y(√y)dx = 3x^2 dx  d(xy(√y))= (3/2)x^2 dx  2xy(√y) = x^3 +c  .

$$\mathrm{3}{x}\sqrt{{y}}{dy}+\mathrm{2}{y}\sqrt{{y}}{dx}\:=\:\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$${d}\left({xy}\sqrt{{y}}\right)=\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} {dx} \\ $$$$\mathrm{2}{xy}\sqrt{{y}}\:=\:{x}^{\mathrm{3}} +{c}\:\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18

((3xdy+2ydx)/(3x×dx))=(x/(√y))  3x(√y) dy+2y^(3/2) dx=3x^2 dx  xd(2y^(3/2) )+2y^(3/2) d(x)=d(x^3 )  d(x×2y^(3/2) )=d(x^3 )  x×2y^(3/2) =x^3 +c

$$\frac{\mathrm{3}{xdy}+\mathrm{2}{ydx}}{\mathrm{3}{x}×{dx}}=\frac{{x}}{\sqrt{{y}}} \\ $$$$\mathrm{3}{x}\sqrt{{y}}\:{dy}+\mathrm{2}{y}^{\frac{\mathrm{3}}{\mathrm{2}}} {dx}=\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$${xd}\left(\mathrm{2}{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)+\mathrm{2}{y}^{\frac{\mathrm{3}}{\mathrm{2}}} {d}\left({x}\right)={d}\left({x}^{\mathrm{3}} \right) \\ $$$${d}\left({x}×\mathrm{2}{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)={d}\left({x}^{\mathrm{3}} \right) \\ $$$${x}×\mathrm{2}{y}^{\frac{\mathrm{3}}{\mathrm{2}}} ={x}^{\mathrm{3}} +{c} \\ $$

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