Question

how do I attempt non-homogeneous separable ODEs? dy/dx=-(4x+5y)/(x+2y), I know its not an Exact differential equation

Answer 1

http://i.imgur.com/d0g4ZSF.png?1

Answer 2

There is a very specific technique to this. First let:
\[y = xv(x) \implies y' = v(x) + x v'(x)\]
From there we have:
\[v(x) + x v'(x) = \frac{-(4x + 5xv(x))}{x+2xv(x)}\]
Write v'(x) in a more suggestive way:
\[v(x) + x \frac{d v(x)}{dx} = \frac{-(4 + 5 x v(x))}{x+2xv(x)}\]
Solve for v'(x) in terms of x and v(x) giving:
\[\frac{d v(x)}{dx} = \frac{1}{x} \left[ \frac{-4 - 5v(x)}{1+2v(x)} -v(x)\right]=\frac{-6v(x)-2v(x)^2-4}{x(1+2v(x))}\]
Which it follows that:
\[\huge\boxed{\int\limits \frac{\frac{dv(x)}{dx} (1+2v(x))}{v(x)^2+3v(x)+2} = \int\limits \frac{-2 dx}{x}}\]
Can you finish that?