Question

IVP:
y+x(y+1)dy/dx=0
y(e^2)=1

Answer 1

I just need hints

Answer 2

c'mon. The first thing to always check is if it is separable. Is it?

Answer 3

yes, looks like this i think: xdx=-y/(y+1)dy

Answer 4

double check that

Answer 5

my algebra is shot, i'm sorry I can"t figure out what it is :(

Answer 6

in which case you are deep trouble. Solving differential equations assumes that you are comfortable with basic algebraic manipulation.

Sorry to be blunt, but don't give up so easily. Try again.

Answer 7

It's no problem, I know I'm not where I need to be, I'll give it another go

Answer 8

How about: (y+1)/-y dy= x dx ??? Am I even close?

Answer 9

\[\Huge \checkmark \] Don't mean to interrupt, I was just looking in.

Answer 10

Okay :) does ✓ mean that I finally got it right?

Answer 11

If
y+x(y+1)dy/dx=0

then
x(y+1) . dy/dx = -y

-(y+1)/y . dy/dx = 1/x

-(y+1)/y . dy = 1/x . dx

Answer 12

The problem will be the integration, not the separation, in my opinion.

Answer 13

Oh I see

Answer 14

What confuses me entirely is this part:

\[\Large \int -\frac{y+1}{y}dy= \int \frac{1}{x}dx \\
\Large \int \left(-1 - \frac{1}{y} \right)dy= \int \frac{1}{x}dx\] so, considering that I didn't sleep to much but I end up with \[\Large -y(x) - \ln |y|=\ln|x| + C \]
Can somebody show me where I went wrong?

Answer 15

That's what I got....

Answer 16

yes, looks right. Now apply the initial condition and see what you end up with.

Answer 17

I see, I am used to get the equation in explicit form first, here that would be rather hard it seems.

Answer 18

Yes, it won't be possible write down an expression for y(x) explicitly, but only implicitly.

Answer 19

C=-3 ?

Answer 20

clearly yes

Answer 21

Woot, thanks a lot for your help James, you've been so helpful!

Answer 22

I would write this now as

y + ln y = - ln x -3
or
ln(y.e^y) = ln(1/(e^3.x))

Hence

y.e^y = 1/(e^3.x)

Answer 23

You should now verify this satisfies the IVP by differentiating (implicitly) and checking also double checking the initial value.