Find General Solution of (x + y) y' = 1

[sumrtym77]

I am having some trouble with finding the general solution of (x+y)y'=1. I think to start I know I need to do substitution but will I be starting with dy/dx=1/(x+y)?

 

[Deleted member 4993]

Re: Find General Solution

I am having some trouble with finding the general solution of (x+y)y'=1. I think to start I know I need to do substitution

but will I be starting with dy/dx=1/(x+y)?

What are the methods you have been taught for first order - ODE?

I would start substitution

v = x + y

 

[sumrtym77]

Re: Find General Solution

I tried that. Let u= x+y, so, y=u-x, dy/dx= du/dx - 1. Then, du/dx = (1/u) + 1.

So I have, u du = 1+ 1 dx. S u du = S 2 dx => (1/2) u^2 = 2x + c
u^2 = 4x + 2c
u = sqrt (4x + 2c)
(x + y) = sqrt (4x +2c)
y = sqrt (4x + 2c) - x

That's what I done, but I know the answer is supposed to be y= ln(x + c + 1).

 

[Deleted member 4993]

Re: Find General Solution

I tried that. Let u= x+y, so, y=u-x, dy/dx= du/dx - 1. Then, du/dx = (1/u) + 1.

So I have, u du = 1+ 1 dx. <<< How is that?

It should be
\(\displaystyle \frac{du}{\frac{1}{u}+1} \, = \, dx\)

Now continue....


S u du = S 2 dx => (1/2) u^2 = 2x + c
u^2 = 4x + 2c
u = sqrt (4x + 2c)
(x + y) = sqrt (4x +2c)
y = sqrt (4x + 2c) - x

That's what I done, but I know the answer is supposed to be y= ln(x + c + 1).

 

[sumrtym77]

Re: Find General Solution

Thank you. I got it to come out with your help. I have the biggest problems just getting started.