(x^2 + 4y^2+4xy)dy=(2x+4y+1)dx
How do I solve the xy part? Where will it go?
I hav tried substitution.. But its of no use..
What I noticed first is that the DE can be written as:
$\displaystyle (x+2y)^2dy=[2(x+2y)+1]dx$
So the two sides have something in common, but I'm not entirely sure if you can do something with this.
Excellent! And, yes, you can make use of that. Let u= x+2y so that du= dx+ 2dy or dy= (1/2)du- (1/2)dx. That makes the differential equation $\displaystyle u^2((1/2)du- (1/2)dx)= (2u+1)dx$ so $\displaystyle (1/2)u^2du= (2u+ 1/2)dx$ or $\displaystyle u^2du= (4u+1)dx$. That's a separable equation.
$\displaystyle \textbf{x}(u)=u - 2\, \ln\!\left(u + \sqrt{2} + 2\right) - 2\, \ln\!\left(u - \sqrt{2} + 2\right) - \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{2} + \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{2}+C$
$\displaystyle \textbf{y}(u) = \frac{1}{2}(u-x)=\ln\!\left(u + \sqrt{2} + 2\right) + \ln\!\left(u - \sqrt{2} + 2\right) + \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{4} - \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{4}-\frac{C}{2}$