Find a general solution for $y'=\frac{3x+4y+2}{2x+y+3}$ $$\frac{dy}{dx}=\frac{3x+4y+2}{2x+y+3}$$
I tried to solve this problem as following,
like 'exact form problem',
$$(2x+y+3)dy - (3x+4y+2)dx = 0$$
This equation is not exact yet, so i tried to find integrating factor (which make equation into exact form). However, i can't find any factor.
My Question:


*

*How to solve this?

*Is there method using linear transformation ? ( like $3x+4y+2 \rightarrow x$ , $2x+y+3 \rightarrow y$)
Second question is motivated by fact that when we find vloume $(3x+y)^2 + (2x+5) \leq 1$, we can use linear transformation to find vloume by multiplying determinant.
 A: This has the type of a homogeneous ODE. Set $y-b=(x-a)u$ where $a,b$ solve
$$
3a+4b+2=0,\\
2a+b+3=0.
$$
This gives $a=-2$, $b=1$, and
$$
(x+2)u'+u=y'=\frac{3(x+2)+4(y-1)}{2(x+2)+(y-1)}=\frac{3+4u}{2+u},
$$
which is now separable.

Or introduce another parameter on the solution curves so that the equation separates into a linear system
$$
\pmatrix{\dot x(t)\\\dot y(t)}
=
\pmatrix{2&1\\3&4}
\pmatrix{x(t)\\y(t)}
+
\pmatrix{3\\2}
$$
with characteristic equation $0=λ^2-6λ+5=(λ-5)(λ-1)$. Left eigenvectors are $(1,1)$ and $(3,-1)$, giving the decoupled equations
$$
(\dot x+\dot y)=5(x+y)+5\\
(3\dot x-\dot y)=(3x-y)+7
$$
which are easy to solve. Then eliminate $t$
A: Let the integrating factor $f$ depend only on $x.$ Then it should satisfy the equation $$2f+2xf'+yf'+3f'=-4f.$$ This simplifies to make $$\frac {f'}{f}=-\frac {6}{2x+y+3},$$ which we may now easily integrate to find a solution $$f=\frac{1}{(3+y+2x)^3}.$$
But this is probably more laborious than the first of the methods suggested by @LutzL above.
