Differential Equations: Finding the General Solution


by amsscorpio
Tags: differential, equations, solution
amsscorpio
amsscorpio is offline
#1
Sep2-09, 07:31 PM
P: 3
Hello, this is the first time I post here, I'm really stumped and tried everything, even my TI-89 calculator won't give me something nice XD

1. The problem statement, all variables and given/known data

Find the general solution of

dy/dt= 1/(ty+t+y+1)


2. Relevant equations
No relevant equations.


3. The attempt at a solution

The first step I did was,

(ty+t+y+1)dy = 1dt

by cross multiplying proportions.
I can't figure out anyway to seperate my terms to the proper places...I then tried this:

integral( ty+t+y+1 dy ) = integral( 1dt )
and resulted in:

ty^2/2 + ty + y^/2 + y + c = t

But this leads me nowhere...Any ideas? Or should I come into conclusion that this differential equation is not valid and unable to do?
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amsscorpio
amsscorpio is offline
#2
Sep2-09, 07:48 PM
P: 3
Ahh I THOUGHT I tried everything XD simple algebra error...

I found that (ty + t + y + 1) = (t + 1)(y + 1).

when factored...
then my next step would be

(y + 1) dy = dt/(t + 1).
which then i can integrate both sides to give me

y^2/2 + y + c = ln|t+1| + c

solving for y ultimately gives me

y = + or - (sqrt( 4ln|t+1| +1 ) +1 )/2

All over 2.

Am I right?
rock.freak667
rock.freak667 is offline
#3
Sep2-09, 08:00 PM
HW Helper
P: 6,214
Quote Quote by amsscorpio View Post
Ahh I THOUGHT I tried everything XD simple algebra error...

I found that (ty + t + y + 1) = (t + 1)(y + 1).

when factored...
then my next step would be

(y + 1) dy = dt/(t + 1).
which then i can integrate both sides to give me

y^2/2 + y + c = ln|t+1| + c

solving for y ultimately gives me

y = + or - (sqrt( 4ln|t+1| +1 ) +1 )/2

All over 2.

Am I right?
I replied in your other thread that same hint but it looks correct.

But just know that when you have things like y2+y3=ln(x2+x+1)+e87x+x3+C

you don't always need to make 'y' the subject of the formula, you can leave it as is.

Mark44
Mark44 is offline
#4
Sep2-09, 08:02 PM
Mentor
P: 21,067

Differential Equations: Finding the General Solution


Quote Quote by amsscorpio View Post
Ahh I THOUGHT I tried everything XD simple algebra error...

I found that (ty + t + y + 1) = (t + 1)(y + 1).

when factored...
then my next step would be

(y + 1) dy = dt/(t + 1).
which then i can integrate both sides to give me

y^2/2 + y + c = ln|t+1| + c
You don't need constants on both sides in the equation above.
Quote Quote by amsscorpio View Post
solving for y ultimately gives me

y = + or - (sqrt( 4ln|t+1| +1 ) +1 )/2
Now you don't have any constants. The constant that should have been on the right side in your previous equation will show up in this one inside the radical.
Quote Quote by amsscorpio View Post
All over 2.

Am I right?
You can check by differenting and seeing whether it is a solution of the original differential equation. You might want to make life a bit easier on yourself by working with y2 rather than y, differentiating implicitly.
amsscorpio
amsscorpio is offline
#5
Sep2-09, 08:03 PM
P: 3
Ahh I don't know how to choose best answer + feedback on yahoo answers lol, I did it to have more chances of geting help, thank you and yes I will remember that.


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