# Nonseparable Differential Equation

1. Feb 29, 2008

### altcmdesc

I know that this differential equation is not separable, but is there a way to solve it?

dy/dx=y+x

I've tried a substitution of y=vx:

(dv/dx)x+v=x+vx
(dv/dx)x=x+vx-v
dv/dx=1+v-(v/x)

I'm stuck trying to rewrite that as a product of v and x.

Any help is appreciated!

2. Feb 29, 2008

### arildno

Set u=y+x,

then you get:
du/dx=dy/dx+1,

du/dx=1+u, which IS separable.

Alternatively, find an integrating factor to your diff.eq.

3. Feb 29, 2008

### altcmdesc

By using that method, my answer was: eˣ-x-1.

Correct?

Thanks!

4. Feb 29, 2008

### John Creighto

Of course there are many methods to solve the above equations as they are a system of linear differential equations. The methods solved above are great for their simplicity but not so great in terms of generality.

5. Feb 29, 2008

### arildno

Why not check it out?

We have: y=e^x-x-1,

whereby:

dy/dx=e^x-1=(e^x-x-1)+x=y+x

so that is indeed A solution.

You still lack the general solution.

6. Mar 2, 2008

### AhmedEzz

isn't the general solution, the same thing but with the constant not being defined?

if not then please clarify it for me.

7. Mar 2, 2008

### rock.freak667

Yes it is...but your solution was $y=e^x-x-1$ which is a particular solution. the general sol'n would be $y=e^x-x-1+C$....C= a constant. Never forget the constant of integration.

8. Mar 2, 2008

### coomast

The general solution is:

$$y(x)= A\cdot e^x -x- 1$$

If you set A=1 then you get the particular solution of altcmdesc. However, the general solution is also obtained via the method of Arildno.

Nothing to do with adding a constant just like that, rock.freak667, you have to add it at the right place.