no info.this is from diff eqtn
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This can be written:

y'=½(1+1/(2x+2y+1)) and y=x/2+½∫dx/(2x+2y+1).

Consider y'=1/(2x+2y+1), or y'(2x+2y+1)=1, and consider d(2x+2y+1)/dx=2+2y'.

As an experiment let y=aln(2x+2y+a), then y'=a(2+2y')/(2x+2y+b).

We can rewrite this: y'(2x+2y+b)=a(2+2y') or y'(2x+2y+b-2a)=2a.

Therefore if b-2a=1 and 2a=1, a=½, b=2, we have y'(2x+2y+1)=1, which is what we are trying to solve!

But this is only part of the answer, so let’s assume we can write a general answer in the form:

y=Ax+Bln(2x+2y+C)+k where k is the constant of integration. We need to find coefficients A, B, C if we can.

y'=A+B(2+2y')/(2x+2y+C).

y'(2x+2y+C)=A(2x+2y+C)+2B+2By'.

Rewriting: y'(2x+2y+C-2B)=2Ax+2Ay+AC+2B=x+y+1.

Equating coefficients, A=½, AC+2B=1, so ½C+2B=1, C-2B=1, from which 3C/2=2 and C=4/3, making B=⅙.

Therefore the solution is y=½x+⅙ln(2x+2y+4/3)+k.

by Top Rated User (804k points)
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Here is an alternative solution:

dy/dx=(x+y+1)/(2x+2y+1)

Let v=x+y, dv/dx=1+dy/dx=1+(v+1)/(2v+1)=(3v+2)/(2v+1).

∫dx=∫((2v+1)/(3v+2))dv.

Let u=3v+2, du=3dv, dv=du/3, v=(u-2)/3.

∫dx=∫((2u-4)/3+1)/u)du/3=(2/9)∫((2u-1)/u)du.

∫dx=(1/9)∫(2-1/u)du=2u/9-(1/9)ln|u|.

9∫dx=2u-ln|u|, 9x+C=2u-ln|u|.

C=-3x+6y-ln|3x+3y+2| where C has been adjusted to absorb the constant 4.

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