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Basic Diffy Q

  • #1

Homework Statement


[itex]dy/dx = 1/(x+y)[/itex]


Homework Equations


Errr. None that I know of.


The Attempt at a Solution


[itex]
v=x+y
[/itex]

[itex]
dy/dx=1/v
[/itex]

[itex]
dv/dx=1+dy/dx
[/itex]

[itex]
dv/dx=1+1/v
[/itex]

[itex]
dv/dx=(v+1)/v
[/itex]

[itex]
dv*v/(v+1)=dx
[/itex]

[itex]
v+1/(v+1) - 1/(v+1) = v/(v+1)
[/itex]

[itex]
\int 1\,dv - \int 1/(v+1)\,dv=\int 1\,dx
[/itex]

[itex]
v - \log (v+1) = x + C
[/itex]

[itex]
e^v/(v+1)=Ce^x
[/itex]

[itex]
e^(x+y)/(x+y+1)=Ce^x
[/itex]


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

Answers and Replies

  • #2
33,075
4,779

Homework Statement


[itex]dy/dx = 1/(x+y)[/itex]


Homework Equations


Errr. None that I know of.


The Attempt at a Solution


[itex]
v=x+y
[/itex]

[itex]
dy/dx=1/v
[/itex]

[itex]
dv/dx=1+dy/dx
[/itex]

[itex]
dv/dx=1+1/v
[/itex]

[itex]
dv/dx=(v+1)/v
[/itex]

[itex]
dv*v/(v+1)=dx
[/itex]

[itex]
v+1/(v+1) - 1/(v+1) = v/(v+1)
[/itex]

[itex]
\int 1\,dv - \int 1/(v+1)\,dv=\int 1\,dx
[/itex]

[itex]
v - \log (v+1) = x + C
[/itex]

[itex]
e^v/(v+1)=Ce^x
[/itex]

[itex]
e^(x+y)/(x+y+1)=Ce^x
[/itex]


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.
 

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