Differential equation, determinant = 0

Question

$$(x+2y+1)y' = 2x+4y+3$$

So i've tried to write it like:

$$y' = \frac{(2x+4y+3)}{x+2y+1}$$

and when i do the determinant i get that its $0$.

I've tried to write it then with $\alpha$ and $\beta$ like: $$2\alpha + 4\beta +1 = 0$$ $$\alpha + 2\beta + 1 = 0$$

But i also end up with $0$ no mather what i try. Im not sure what other way should I try. Can you help me out?

Answer 1

In that case you can cancel to $$ y'=2+\frac1{x+2y+1}. $$ You could also substitute $u=x+2y$ to find $$ u'=1+2y'=5+\frac2{u+1} $$

Answer 2

To get rid of the denominator you introduced, define $$x+2y+1=z\implies y=\frac{1}{2} (z-x-1)\implies y'=\frac{1}{2} \left(z'-1\right)$$ Replace in the original equation $$(x+2y+1)y' = 2x+4y+3$$ and you will find a nice separable equation.