dy/dx=(2x+1)/(3y+x+2). To convert non-homogeneous DE to a homogeneous DE, let x=X+h, y=Y+k.
Therefore dy/dx=dY/dX, and:
dY/dX=(2(X+h)+1)/(3(Y+k)+X+h+2),
dY/dX=(2X+2h+1)/(3Y+3k+X+h+2).
We need to find values for h and k such that the constant terms disappear.
2h+1=0⇒h=-½; 3k+h+2=0=3k-½+2⇒k=-½.
dY/dX=2X/(3Y+X). This is now homogeneous so let Y=vX, where v is a function of X.
dY/dX=v+dv/dX=2X/(3vX+X)=2/(3v+1),
dv/dX=2/(3v+1)-v=(2-3v2-v)/(3v+1)=(2-3v)(1+v)/(3v+1).
We can now separate the variables X and v and integrate:
∫{(3v+1)/[(2-3v)(1+v)]}dv=∫dX+C where C is the integration constant.
We can split the first integral into partial fractions:
(3v+1)/[(2-3v)(1+v)]=A/(2-3v)+B/(1+v) where A and B are constants to be found.
3v+1=A+Av+2B-3Bv;
A+2B=1, A-3B=3⇒1-2B-3B=3, 1-5B=3, 5B=-2, B=-⅖; A=1-2B=9/5.
The integral becomes ∫[9/(10-15v)-2/(5+5v)]dv=X+C,
-⅗ln|10-15v|-⅖ln|5+5v|=X+C, substituting for v, X and Y:
v=Y/X=(y+½)/(x+½)=(2y+1)/(2x+1), the solution becomes:
-⅗ln|10-15(2y+1)/(2x+1)|-⅖ln|5+5(2y+1)/(2x+1)|=x+C1, C1=C+½,
3ln|(2x+1)/(20x-30y-5)|+2ln|(2x+1)/(10x+10y+10)|=5x+C2, where C2=5C1,
3ln|(2x+1)/(4x-6y-1)|+2ln|(2x+1)/(x+y+1)|=5x+C3, C3=C2+ln(53×102)=C2+ln(12500),
ln|(2x+1)5/[(4x-6y-1)3(x+y+1)2]|=5x+C3, which can be expressed as an exponential:
(2x+1)5/[(4x-6y-1)3(x+y+1)2]=e5x+c, where c is some constant.
PARTIAL CONFIRMATION OF SOLUTION
Start with -⅗ln|10-15v|-⅖ln|5+5v|=X+C, differentiating wrt X:
(-⅗)(-15dv/dX)/(10-15v)-(⅖)(5dv/dX)/(5+5v)=1,
9dv/dX/(10-15v)-2dv/dX/(5+5v)=1,
⅕(dv/dX)(9/(2-3v)-2/(1+v))=1,
⅕(dv/dX)(9+9v-4+6v)/(2-v-3v2)=1,
⅕(dv/dX)(5+15v)/(2-v-3v2)=1,
(dv/dX)(1+3v)/(2-v-3v2)=1,
dv/dX=(2-v-3v2)/(1+3v)=2/(3v+1)-v,
v+dv/dX=2/(3v+1). Now we can back-substitute v, X and Y to get a DE in x and y only:
Y=vX, dY/dX=v+dv/dX⇒dY/dX=2/(3Y/X+1)=2X/(3Y+X).
X=x+½, Y=y+½, dY/dX=dy/dx=y'=2(x+½)/(3(y+½)+x+½)=(2x+1)/(3y+x+2).
This confirms the solution from the starting point -⅗ln|10-15v|-⅖ln|5+5v|=X+C.