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

[alexmahone]

Solve \(\displaystyle y'(x)=5x^3(y-1)^\frac{1}{5}\), \(\displaystyle y(0)=1\).

I found 3 solutions: \(\displaystyle y=x^5+1\), \(\displaystyle y=-x^5+1\) and the constant solution \(\displaystyle y=1\).

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

 

[I like Serena]

Solve \(\displaystyle y'(x)=5x^3(y-1)^\frac{1}{5}\), \(\displaystyle y(0)=1\).

I found 3 solutions: \(\displaystyle y=x^5+1\), \(\displaystyle y=-x^5+1\) and the constant solution \(\displaystyle y=1\).

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 $y=1$.

 

[alexmahone]

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

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

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

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

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

6) \(\displaystyle \displaystyle y =\begin{cases} 1\ \text{if}\ x \ge 0\\ -x^5+1\ \text{if}\ x< 0\end{cases}\)
-
I get only 6, not infinitely many!

 

[I like Serena]

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?

 

[alexmahone]

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 $x=a$.

 

[I like Serena]

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.

 

[I like Serena]

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.