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Question Number 89243 by M±th+et£s last updated on 16/Apr/20

solve the following diffirntial equation  1)(2x+y)dx+(x+y)dy=0  2)(3x−y)dx−(x−y)dy=0  3) (cos(x)+y)dx + (2y+x)dy=0

$${solve}\:{the}\:{following}\:{diffirntial}\:{equation} \\ $$$$\left.\mathrm{1}\right)\left(\mathrm{2}{x}+{y}\right){dx}+\left({x}+{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{3}{x}−{y}\right){dx}−\left({x}−{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left({cos}\left({x}\right)+{y}\right){dx}\:+\:\left(\mathrm{2}{y}+{x}\right){dy}=\mathrm{0} \\ $$

Answered by TANMAY PANACEA. last updated on 16/Apr/20

1)2xdx+ydx+xdy+ydy=0  d(x^2 )+d(xy)+(1/2)d(y^2 )=dc  d(x^2 +xy+(y^2 /2))=dc  x^2 +xy+(y^2 /2)=c

$$\left.\mathrm{1}\right)\mathrm{2}{xdx}+{ydx}+{xdy}+{ydy}=\mathrm{0} \\ $$$${d}\left({x}^{\mathrm{2}} \right)+{d}\left({xy}\right)+\frac{\mathrm{1}}{\mathrm{2}}{d}\left({y}^{\mathrm{2}} \right)={dc} \\ $$$${d}\left({x}^{\mathrm{2}} +{xy}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)={dc} \\ $$$${x}^{\mathrm{2}} +{xy}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}={c} \\ $$

Answered by TANMAY PANACEA. last updated on 16/Apr/20

2)3xdx−ydx−xdy+ydy=0  (3/2)d(x^2 )−d(xy)+(1/2)d(y^2 )=dc  d(((3x^2 )/2)−xy+(y^2 /2))=dc  3x^2 −2xy+y^2 =2c

$$\left.\mathrm{2}\right)\mathrm{3}{xdx}−{ydx}−{xdy}+{ydy}=\mathrm{0} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{d}\left({x}^{\mathrm{2}} \right)−{d}\left({xy}\right)+\frac{\mathrm{1}}{\mathrm{2}}{d}\left({y}^{\mathrm{2}} \right)={dc} \\ $$$${d}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{xy}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)={dc} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}+{y}^{\mathrm{2}} =\mathrm{2}{c} \\ $$

Answered by TANMAY PANACEA. last updated on 16/Apr/20

cosxdx+ydx+xdy+2ydy=0  d(sinx)+d(xy)+d(y^2 )=dc  d(sinx+xy+y^2 )=dc  sinx+xy+y^2 =c

$${cosxdx}+{ydx}+{xdy}+\mathrm{2}{ydy}=\mathrm{0} \\ $$$${d}\left({sinx}\right)+{d}\left({xy}\right)+{d}\left({y}^{\mathrm{2}} \right)={dc} \\ $$$${d}\left({sinx}+{xy}+{y}^{\mathrm{2}} \right)={dc} \\ $$$${sinx}+{xy}+{y}^{\mathrm{2}} ={c} \\ $$

Commented by M±th+et£s last updated on 16/Apr/20

hod bless you sir .

$${hod}\:{bless}\:{you}\:{sir}\:. \\ $$

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