Verified answer

first divide by x

dy/dx + 2y/x = 3/x

Now multiply the whole equation by e^∫2/x

so multiply by e^( Ln( x^2 ))

= x^2

x^2 dy/dx + 2x y = 3x

now the left hand side of the equation becomes

d/dx ( y x^2 ) = 3x

integrate both sides

y x^2 = (3/2) x^2 + C

now divide everything by x^2

y = ( 3/2 ) + C/x^2

Differntials aren'tmy specialty, but I'll give it a try.

First I'm going to separate the functions to try and get the dy/dx on one side, and everything that isn't directly attached to it on the other side. Then I'm going to try and put the y and dy together and the x and dx together so I can integrate them.

x(dy/dx)+2y=3

x(dy/dx)=3-2y

x/dx=(3-2y)/dy

Now I'm going essentially flip both sides to get the dy and dx both on the top.

1/x(dx)=1/(3-2y)(dy)

When we integrate we end up with-

ln(x)+c=-2ln(3-2y) or ln(x)+C=ln[1/(3-2y)^2]

Next we are going get rid of the logs by making both sides a power of e.

e^[ln(x)+C]=e^ln[1/(3-2y)^2]

On the left side we are going to make e^[ln(x)+C] into this e^ln(x)*e^C which simplifies to xe^C or Cx

So now we have-

Cx=1/(3-2y)^2

The rest is basically solving for y.

I'm going to start by flipping both sides again so that our y is on the top.

1/(Cx)=(3-2y)^2

square root both sides. subract 3, divide by -2.

sqrt[1/(Cx)]=3-2y

sqrt[1/(Cx)]-3=-2y

sqrt[1/(Cx)]/(-2)+3/2=y

I think you can fudge the -2 into the Constant making it sqrt[1/(Cx)]+3/2=y

But I think it's not a good idea since I'm rusty on this stuff.