How Can You Solve the Differential Equation dy/dx = 1/(x+y)?

[domesticbark]

Homework Statement

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

Homework Equations

Errr. None that I know of.

The Attempt at a Solution

$v=x+y$

$dy/dx=1/v$

$dv/dx=1+dy/dx$

$dv/dx=1+1/v$

$dv/dx=(v+1)/v$

$dv*v/(v+1)=dx$

$v+1/(v+1) - 1/(v+1) = v/(v+1)$

$\int 1\,dv - \int 1/(v+1)\,dv=\int 1\,dx$

$v - \log (v+1) = x + C$

$e^v/(v+1)=Ce^x$

$e^(x+y)/(x+y+1)=Ce^x$


I'm just wondering whether I can simplify this or maybe solve it another way so I can get y= (stuff)

 

[Mark44]

Homework Statement

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

Homework Equations

Errr. None that I know of.

The Attempt at a Solution

$v=x+y$

$dy/dx=1/v$

$dv/dx=1+dy/dx$

$dv/dx=1+1/v$

$dv/dx=(v+1)/v$

$dv*v/(v+1)=dx$

$v+1/(v+1) - 1/(v+1) = v/(v+1)$

$\int 1\,dv - \int 1/(v+1)\,dv=\int 1\,dx$

$v - \log (v+1) = x + C$

$e^v/(v+1)=Ce^x$

$e^(x+y)/(x+y+1)=Ce^x$


I'm just wondering whether I can simplify this or maybe solve it another way so I can get y= (stuff)

Assuming your work is correct (I didn't check it), you can leave the solution in implicit form. You might not be able to solve the equation you ended up with for y as an explicit function of x.