can somebody solve (2x+y+3)dy =(x+2y+3)dx?

Question

solve differential equation

Answer 1

Um ok.
First rearrange the equation.
So you get: dy/dx = (x+2y+3)/(2x+y+3)
Now, this is just an implicit differential.
Just find the derivative of (x+2y+3)/(2x+y+3).

Answer 2

Some of the above answers are not working when you put them back into the differential equation.
Separation of variables is difficult because it's not easy to separate them. Just look at the equation and see.

But you will notice a degree of symmetry and linearity.
So assume a solution of the form
y = ax + b
dy/dx = a = (x+2y+3) / (2x+y+3)
= [x*(1+2a) + (2b+3)] / [x*(2+a) + (b+3)]

Since this must be true for all x, then we can conclude that
1 + 2a = a*(2+a)
which yields a = -1 or +1.
And 2b+3 = a*(b+3)
which yields b = -2 or 0

So two possible solutions are:
y = x
y = - x - 2

**EDIT**
ksoileau, below, is da man. Great solution.

Answer 3

Can you write the question again? I don't get it is it x^(3dy/dx)+2x^(2y)=5 or x^3 dy/dx + 2x^2 y=5

Answer 4

Solve by using 'Separation of Variables'.

-y+3 dy = -x + 3 dx

Integration yields:

-0.5y^2 +3y = -0.5x^2 +3x

Answer 5

I think you must integrate each part of the egality, you obtein

(2x.y + y²/2 + 3.y) +C = (x²/2 + 2.y.x + 3.x) + C' where c and c' are real
ie
0.5 ( y² - x²) = 3 (x-y) +k where k is real
ie
( y - x ) ( y + x ) = 6 ( x - y ) k

y + x = K-6

y = -x +k' k' is real

Answer 6

2xy+ y^2/2 +3y=x^2/2 + 2yx+3x

Answer 7

integrate both sides. Then use algebra to solve.

Answer 8

-1/2*(y-x)^(-2)
-(1+x)*(y-x)^(-3)
=C

Set u=y-x
Then (2x+u+x+3)(du+dx) =(x+2(u+x)+3)dx
(3x+u+3)(du+dx) =(3x+2u+3)dx
(u+3)du =udx-3xdu
u^(-4)(u+3)du =u^(-4)udx-3xu^(-4)du
(u^(-3)+3u^(-4))du =u^(-3)dx-3xu^(-4)du
-1/2*u^(-2)-u^(-3) =xu^(-3)+C
-1/2*(y-x)^(-2)-(y-x)^(-3) =x(y-x)^(-3)+C

Edit: Thanks, Dr D!