Differential Equations: Finding the General Solution

In summary, the person is trying to solve a differential equation, but gets stuck because they can't find the constants for the equation. They eventually find that y = + or - (sqrt( 4ln|t+1| +1 ) +1 )/2, and all over 2.
  • #1
amsscorpio
3
0
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 separate 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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  • #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
amsscorpio said:
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
amsscorpio said:
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.
amsscorpio said:
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.
amsscorpio said:
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.
 

1. What is a differential equation?

A differential equation is an equation that contains one or more derivatives of an unknown function. It relates the function to its derivatives and can be used to model various phenomena in science and engineering.

2. What is the general solution of a differential equation?

The general solution of a differential equation is a family of functions that satisfies the equation. It contains all possible solutions and can be obtained by solving the equation without specifying initial conditions.

3. How do you find the general solution of a differential equation?

To find the general solution of a differential equation, you need to solve the equation by using various methods such as separation of variables, substitution, or integrating factors. This will give you a general solution in the form of a function with arbitrary constants.

4. What are initial conditions in a differential equation?

Initial conditions are values that are used to uniquely determine a particular solution of a differential equation from the general solution. They are typically given at a specific point or boundary of the function.

5. Why is finding the general solution of a differential equation important?

Finding the general solution of a differential equation is important because it allows us to find a wide range of solutions to a given problem. This is especially useful in modeling real-world phenomena, where there may be multiple possible solutions that can be described by a single differential equation.

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