Solve (x^2 + 4y^2+4xy)dy=(2x+4y+1)dx

**animesh271094** · August 14th 2011, 07:21 AM

(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..

**Siron** · August 14th 2011, 08:31 AM

What I noticed first is that the DE can be written as:
$(x+2y)^2dy=[2(x+2y)+1]dx$

So the two sides have something in common, but I'm not entirely sure if you can do something with this.

**HallsofIvy** · August 14th 2011, 12:54 PM

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 $u^2((1/2)du- (1/2)dx)= (2u+1)dx$ so $(1/2)u^2du= (2u+ 1/2)dx$ or $u^2du= (4u+1)dx$ . That's a separable equation.

**zzzoak** · August 14th 2011, 03:16 PM

Sorry I get separable

$u^2du= (u^2+4u+2)dx \; .$

**SammyS** · August 14th 2011, 07:36 PM

Originally Posted by HallsofIvy

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 $u^2((1/2)du- (1/2)dx)= (2u+1)dx$ so $(1/2)u^2du= (2u+ 1/2)dx$ or $u^2du= (4u+1)dx$ . That's a separable equation.

From $u^2((1/2)du- (1/2)dx)=(2u+1)dx$ , I get:

$u^2du- u^2dx=(4u+2)dx$

$u^2du=(u^2+4u+2)dx$

Which is certainly separable.

**animesh271094** · August 14th 2011, 07:48 PM

Thanx a lot..! i just wanted to get rid of the 'xy' term.. !

**math2009** · August 16th 2011, 01:37 AM

$\textbf{x}(u)=u - 2\, \ln\!\left(u + \sqrt{2} + 2\right) - 2\, \ln\!\left(u - \sqrt{2} + 2\right) - \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{2} + \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{2}+C$

$\textbf{y}(u) = \frac{1}{2}(u-x)=\ln\!\left(u + \sqrt{2} + 2\right) + \ln\!\left(u - \sqrt{2} + 2\right) + \frac{3\, \sqrt{2}\, \ln\!\left(u + \sqrt{2} + 2\right)}{4} - \frac{3\, \sqrt{2}\, \ln\!\left(u - \sqrt{2} + 2\right)}{4}-\frac{C}{2}$

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