Question

pls solve correctly. I will mark you as Brainliest.​

Answer 1

Answer:

Here is a method from which you can solve such equation:

Here is a method from which you can solve such equation:Multiplying by 2x/x^2

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinates

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C√( x^2+y^2)+2 √( 1+tan^2(θ))=C

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C√( x^2+y^2)+2 √( 1+tan^2(θ))=C√( x^2+y^2)+2 √( 1+(y^2/x^2))=C

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C√( x^2+y^2)+2 √( 1+tan^2(θ))=C√( x^2+y^2)+2 √( 1+(y^2/x^2))=C(x+2) √(x^2+y^2)=Cx

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C√( x^2+y^2)+2 √( 1+tan^2(θ))=C√( x^2+y^2)+2 √( 1+(y^2/x^2))=C(x+2) √(x^2+y^2)=CxHope this helps you. Thank you.

Here is a method from which you can solve such equation:Multiplying by 2x/x^22(xdx+ydy)+4y(xdy−ydx)/x^2=0d(x^2+y^2)+4yd(y/x)=0Converting to polar co ordinatesd(r^2)+4rsinθd(tanθ)=02rdr+4rsinθsec^2(θ)dθ=02rdr+4rsinθsec^2(θ)dθ=0dr+2secθtanθdθ=0Integrating,r+2secθ=C√( x^2+y^2)+2 √( 1+tan^2(θ))=C√( x^2+y^2)+2 √( 1+(y^2/x^2))=C(x+2) √(x^2+y^2)=CxHope this helps you. Thank you.All the best!!