Help Solving Diff. Eq. Integration: y^3 = 1/2y^2+t+C

In summary, the conversation is about a user seeking help with their integration of a simple, separable differential equation. They have encountered some trouble with their integration and are looking for guidance on where they may have gone wrong. Another user suggests solving for the constant using the initial condition, and the original user realizes their mistake and thanks them for their help.
  • #1
Tom1
24
0
I am working on a diff. eq. assignment, and have encountered the simple, seperable differential equation:

dy/dt = y^3 with the initial condition y(0)=1

I am posting in this forum, because I feel my trouble is with my integration.

When seperated, I am integrating this equation as [tex]\int1/y^3[/tex]=[tex]\int1dt[/tex].

After doing so, I end up with: -1/2y^2=t+C

After solving for y, I come up with: y=[tex]\sqrt{1/2t}[/tex]+C

The only problem with this is I cannot solve for C, since the initial condition would require division by zero.

Anyone have an clue as to where I went wrong?
 
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  • #2
Just solve for the constant here: -1/2y^2=t+C.
 
  • #3
If I do it then, I end up with C=-1/2y^2-1 which when plugged back into the general solution, results in y disappearing.

Can't be right..
 
  • #4
Tom1 said:
If I do it then, I end up with C=-1/2y^2-1 which when plugged back into the general solution, results in y disappearing.

Can't be right..
Look, you are not followint Mystic998's point and suggestion at all!
YOu have the initial condition y(0)=1, right?
so as Mystic998 suggested

-1/2y^2=t+C.=> -1/2 *1^2=c=> c=-1/2
and remember y(0)=1, means the value of y when t=0
 
  • #5
I'm an idiot, thank you.
 
  • #6
Tom1 said:
I'm an idiot, thank you.
Oh no, i am sure you are not, it happens to all of us sometimes not to notice these minor things!
 

Related to Help Solving Diff. Eq. Integration: y^3 = 1/2y^2+t+C

1. What is the process for solving this differential equation?

The first step in solving this differential equation is to rearrange it into the standard form: dy/dx = f(x,y). From there, you can use the appropriate integration techniques to find the general solution.

2. How do I deal with the constant of integration?

When solving a differential equation, the constant of integration (C) will typically appear in the final answer. It represents all possible solutions to the equation, and can be determined by using initial conditions or boundary values, if given.

3. What is the role of the initial conditions in solving a differential equation?

The initial conditions are values given for the dependent variable (y) and independent variable (x) at a specific point in the domain. These conditions are used to find the particular solution to the differential equation, by plugging them into the general solution and solving for the constant of integration.

4. Can I use any integration technique to solve this differential equation?

While there are many integration techniques, not all will be applicable to every differential equation. Some common techniques include separation of variables, substitution, and integration by parts. It is important to carefully analyze the equation and choose the appropriate technique for solving it.

5. Is there a way to check if my solution to the differential equation is correct?

Yes, you can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. You can also take the derivative of your solution and see if it matches the original equation.

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