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Differential Equations: Finding the General Solution

  • Thread starter amsscorpio
  • Start date
  • #1
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

Homework Statement



Find the general solution of

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


Homework Equations


No relevant equations.


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?
 

Answers and Replies

  • #2
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?
 
  • #3
rock.freak667
Homework Helper
6,230
31
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.
 
  • #4
33,170
4,855
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.
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.
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.
 
  • #5
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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