MHB Can y'(x)=5x^3(y-1)^\frac{1}{5} have infinitely many solutions?

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Solve [math]y'(x)=5x^3(y-1)^\frac{1}{5}[/math], [math]y(0)=1[/math].

I found 3 solutions: [math]y=x^5+1[/math], [math]y=-x^5+1[/math] and the constant solution [math]y=1[/math].

But my prof said there are infinitely many solutions. Which are they?
 
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Alexmahone said:
Solve [math]y'(x)=5x^3(y-1)^\frac{1}{5}[/math], [math]y(0)=1[/math].

I found 3 solutions: [math]y=x^5+1[/math], [math]y=-x^5+1[/math] and the constant solution [math]y=1[/math].

But my prof said there are infinitely many solutions. Which are they?

How about piece wise combinations of those solutions?
And perhaps you can shift one of them to the left or right, combining it with $y=1$.
 
I like Serena said:
How about piece wise combinations of those solutions?
And perhaps you can shift one of them to the left or right, combining it with $y=1$.

Using piecewise combinations I get

1) [math]\displaystyle y =\begin{cases} x^5+1\ \text{if}\ x \ge 0\\ -x^5+1\ \text{if}\ x< 0\end{cases}[/math]

2) [math]\displaystyle y =\begin{cases} -x^5+1\ \text{if}\ x \ge 0\\ x^5+1\ \text{if}\ x< 0\end{cases}[/math]

3) [math]\displaystyle y =\begin{cases} x^5+1\ \text{if}\ x \ge 0\\ 1\ \text{if}\ x< 0\end{cases}[/math]

4) [math]\displaystyle y =\begin{cases} 1\ \text{if}\ x \ge 0\\ x^5+1\ \text{if}\ x< 0\end{cases}[/math]

5) [math]\displaystyle y =\begin{cases} -x^5+1\ \text{if}\ x \ge 0\\ 1\ \text{if}\ x< 0\end{cases}[/math]

6) [math]\displaystyle y =\begin{cases} 1\ \text{if}\ x \ge 0\\ -x^5+1\ \text{if}\ x< 0\end{cases}[/math]
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I get only 6, not infinitely many!
 
Last edited:
How about:
$$ y =\begin{cases} x^5+1 &\text{if}\ x > a\\1 & \text{if}\ x< a\end{cases}$$
where $a$ is an arbitrary real number?
 
I like Serena said:
How about:
$$ y =\begin{cases} x^5+1 &\text{if}\ x > a\\1 & \text{if}\ x< a\end{cases}$$
where $a$ is an arbitrary real number?

But that wouldn't be continuous at $x=a$.
 
Alexmahone said:
But that wouldn't be continuous at $x=a$.

Sorry. You are right.
Let me think about it.
It must be possible to use $y=1$ up to some $x=a$ and from that point switch to a different solution.
 
How about for instance:
$$y=\begin{cases}
1 + (x^4-a^4)^{5/4} &&\text{if } x < a < 0\\
1 &&\text{if } a \le x < b \\
1 - (x^4-b^4)^{5/4} &&\text{if } x \ge b > 0\\
\end{cases}$$It is continuously differentiable. ;)Note that everything around $y=1$ is special, because your solution method breaks down around it.
 

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