Two Solutions for y' = 5x^3(y-1)^1/5 with Initial Condition y(0)=1

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In summary: That is just one type of singular solution. I give some example curves below, A denser set would be more illustrative but I don't have the technology for it. Also cusps not well rendered.Not all cusp loci of de's are solutions but this one is. You can see that the cusp locus does not go through all the solutions. And that they have two branches.(A more familiar type of singular solution, the only one I dabbled with before now, are envelopes, which I think are always solutions of the d.e.). I might go further into this if there is any interest shown, perhaps the OP can add
  • #1
ver_mathstats
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Homework Statement
Find two solutions for the initial condition y(0)=1.
Relevant Equations
Initial condition: y(0)=1
I have an ODE, which is y' = 5x3(y-1)1/5 with the initial condition y(0)=1, I must find two solutions.

My attempt at solving this problem is as follows:

Separate the equation, we get, dy/(y-1)1/5 = 5x3dx.
Integrate both sides, ∫ dy/(y-1)1/5 = ∫ 5x3dx.
We are left with, (5/4)(y-1)4/5 = (5x4)/4 + C.

Now we may solve for C using our initial condition y(0) = 1. Substituting each of the values into (5/4)(y-1)4/5 = (5x4)/4 + C gives us 0 = 0 + C.

We can now isolate for y which gives us y = x20/4 - 1.

Therefore, the two solutions are y = x20/4 - 1 and y = 1 or would it be y = 0?

Is this correct? Did I solve correctly? Thank you.
 
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  • #2
ver_mathstats said:
Homework Statement:: Find two solutions for the initial condition y(0)=1.
Relevant Equations:: Initial condition: y(0)=1

I have an ODE, which is y' = 5x3(y-1)1/5 with the initial condition y(0)=1, I must find two solutions.

My attempt at solving this problem is as follows:

Separate the equation, we get, dy/(y-1)1/5 = 5x3dx.
Integrate both sides, ∫ dy/(y-1)1/5 = ∫ 5x3dx.
We are left with, (5/4)(y-1)4/5 = (5x4)/4 + C.

Now we may solve for C using our initial condition y(0) = 1. Substituting each of the values into (5/4)(y-1)4/5 = (5x4)/4 + C gives us 0 = 0 + C.

We can now isolate for y which gives us y = x20/4 - 1.

Therefore, the two solutions are y = x20/4 - 1 and y = 1 or would it be y = 0?

Is this correct? Did I solve correctly? Thank you.
Yes, this looks fine, and both solutions check.
You could have made life a bit easier by doing some simplification along the way, though.
(5/4)(y-1)4/5 = (5x4)/4 + C can be simplified to (y-1)4/5 = x4 + C', with C' = 4/5 * C.
Also, there's no point in leaving the exponent as 20/4 -- simplify this to just plain 5.
 
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  • #3
Mark44 said:
Yes, this looks fine, and both solutions check.
You could have made life a bit easier by doing some simplification along the way, though.
(5/4)(y-1)4/5 = (5x4)/4 + C can be simplified to (y-1)4/5 = x4 + C', with C' = 4/5 * C.
Also, there's no point in leaving the exponent as 20/4 -- simplify this to just plain 5.
Thank you for the help, yes, but may I ask the second solution is y = 0 right? Just making sure because I was confusing myself with y = 1 and y = 0.
 
  • #4
Sorry, I missed that you had asked a question about whether the second solution was y = 0 or y = 1 -- y = 1 is the correct second solution. It satisfies the DE and the initial condition.
 
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  • #5
Mark44 said:
Sorry, I missed that you had asked a question about whether the second solution was y = 0 or y = 1 -- y = 1 is the correct second solution. It satisfies the DE and the initial condition.
Okay, thank you, I understand now.
 
  • #6
If you have understood the system completely I have only to learn from you. I don't understand the subject and this case completely, but at least a bit more than I did yesterday.

Possibly the way the question was posed may slightly hide something from you. You get two solutions from the given initial condition. but from any other initial condition you only get one. One of the solutions (y = 1) does not involve any arbitrary constant, and is a single curve, not a family, so it is called a 'singular solution'. If you have graphed the function you will have seen that it is the locus of cusps. That is just one type of singular solution. I give some example curves below, A denser set would be more illustrative but I don't have the technology for it. Also cusps not well rendered.

78036BA5-AC3E-4CCA-8597-1882A9F1668F.png


Not all cusp loci of de's are solutions but this one is. You can see that the cusp locus does not go through all the solutions. And that they have two branches.

(A more familiar type of singular solution, the only one I dabbled with before now, are envelopes, which I think are always solutions of the d.e.). I might go further into this if there is any interest shown, perhaps the OP can add considerations or question, but it is open to anyone.
 
Last edited:
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  • #7
epenguin said:
If you have understood the system completely I have only to learn from you. I don't understand the subject and this case completely, but at least a bit more than I did yesterday.

Possibly the way the question was posed may slightly hide something from you. You get two solutions from the given initial condition. but from any other initial condition you only get one. One of the solutions (y = 1) does not involve any arbitrary constant, and is a single curve, not a family, so it is called a 'singular solution'. If you have graphed the function you will have seen that it is the locus of cusps. That is just one type of singular solution. I give some example curves below, A denser set would be more illustrative but I don't have the technology for it. Also cusps not well rendered.

View attachment 276846

Not all cusp loci of de's are solutions but this one is. You can see that the cusp locus does not go through all the solutions. And that they have two branches.

(A more familiar type of singular solution, the only one I dabbled with before now, are envelopes, which I think are always solutions of the d.e.). I might go further into this if there is any interest shown, perhaps the OP can add considerations or question, but it is open to anyone.
I'm still really new to this, have only started learning about it three weeks ago actually! But all of that is really interesting, so far I've just been learning the basics so I'm not too familiar with the term envelopes actually, would love to gain more insight though.
 
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  • #8
ver_mathstats said:
My attempt at solving this problem is as follows:

Separate the equation, we get, dy/(y-1)1/5 = 5x3dx.
Integrate both sides, ∫ dy/(y-1)1/5 = ∫ 5x3dx.
We are left with, (5/4)(y-1)4/5 = (5x4)/4 + C.

Now we may solve for C using our initial condition y(0) = 1. Substituting each of the values into (5/4)(y-1)4/5 = (5x4)/4 + C gives us 0 = 0 + C.

We can now isolate for y which gives us y = x20/4 - 1.
Notice a minor algebra mistake (or typo).

That should be ##y = x^{20/4} + 1 ## or following @Mark44's suggestion ##y = x^{5} + 1 ##.
 
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  • #9
ver_mathstats said:
I'm still really new to this, have only started learning about it three weeks ago actually! But all of that is really interesting, so far I've just been learning the basics so I'm not too familiar with the term envelopes actually, would love to gain more insight though.
Example of what envelopes are #20 here: https://www.physicsforums.com/threa...er-differential-equation.963906/#post-6117815
 
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  • #10
SammyS said:
Notice a minor algebra mistake (or typo).

That should be ##y = x^{20/4} + 1 ## or following @Mark44's suggestion ##y = x^{5} + 1 ##.
Are you sure that's not a mistake? I briefly thought that, decided it was wrong, but now you worry me.
 
  • #11
SammyS said:
Notice a minor algebra mistake (or typo).

That should be ##y = x^{20/4} + 1 ## or following @Mark44's suggestion ##y = x^{5} + 1 ##.
Yes, sorry about that! I noticed afterwards and fixed it. Thank you.
 
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  • #12
epenguin said:
Are you sure that's not a mistake? I briefly thought that, decided it was wrong, but now you worry me.
It was a typo from my own notes! Sorry about that. My writing is messy.
 

Related to Two Solutions for y' = 5x^3(y-1)^1/5 with Initial Condition y(0)=1

1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how one or more variables change over time. It involves derivatives, or rates of change, of the variables in the equation.

2. Why do we need to find solutions for ODEs?

ODEs are used to model real-world phenomena in fields such as physics, engineering, and biology. By finding solutions to these equations, we can gain a better understanding of how these systems behave and make predictions about their future behavior.

3. How do you find solutions for an ODE?

There are several methods for finding solutions to ODEs, including analytical methods such as separation of variables and numerical methods such as Euler's method. The specific method used will depend on the complexity of the equation and the desired level of accuracy.

4. What are the two types of solutions for an ODE?

The two types of solutions for an ODE are general solutions and particular solutions. A general solution contains arbitrary constants and can represent an infinite number of specific solutions. A particular solution is a specific solution with values assigned to the arbitrary constants.

5. Can an ODE have more than two solutions?

Yes, an ODE can have an infinite number of solutions. This is because an ODE is a mathematical equation, and for every set of values assigned to the variables, there can be a unique solution. However, in some cases, an ODE may have a finite number of solutions or no solutions at all.

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