Differential equation uniqueness

[nysnacc]

Homework Statement

Homework Equations

Leibniz notation: dy/dx = f(x) g(y)
integral 1/g(y) dy = integral f(x) dx

The Attempt at a Solution

integral 1/y dy = integral sqrt (abs x) dx

ln (y) = ? because sqrt (abs x) is not integrable at x =0
Then my thought is that y=0 is not unique

 

[member 587159]

Why don't you continue to solve the equation using change of variables?

$\frac{dy}{dx}= y \sqrt{|x|} \Rightarrow \frac{dy}{y} = \sqrt{|x|}dx \Rightarrow \int \frac{dy}{y} = \int\sqrt{|x|}dx \Rightarrow log_e y = \int\sqrt{|x|}dx$

Now, $\int\sqrt{|x|}dx = \dots$

If you have trouble integrating this function due to the absolute value, use the definition of absolute value to integrate it for $x \leq 0$ and/or $x \geq 0$. Then, use some algebra to find the function $y$.

The Attempt at a Solution

integral 1/y dy = integral sqrt (abs x) dx

ln (y) = ? because sqrt (abs x) is not integrable at x =0
Then my thought is that y=0 is not unique

Can you say why you think that?

 

[nysnacc]

So I have to consider two cases, x>0 and x<= 0?
I really don't know how to integrate sqrt (abs x)
:(

 

[member 587159]

So I have to consider two cases, x>0 and x<= 0?
I really don't know how to integrate sqrt (abs x)
:(

Yes, split it up in apart cases (which is not nessecary, but is easier)

$\sqrt{|x|} = \sqrt{x}$ if $x \geq 0$
$\sqrt{|x|} = \sqrt{-x}$ if $x \leq 0$

Those should be standard to integrate.

 

[LCKurtz]

So I have to consider two cases, x>0 and x<= 0?
I really don't know how to integrate sqrt (abs x)
:(

Well, surely if $x>0$ so $|x| = x$ you can integrate $\sqrt{|x|}$, right? As another hint, if the case $x<0$ confuses you, you might use the fact that if $f(x)$ is an even function (which $\sqrt{|x|}$ is), then$$
H(x) = \int_0^x f(u)~du$$ is an odd function. Draw a graph using that and see what it looks like.

 

[nysnacc]

Well, surely if $x>0$ so $|x| = x$ you can integrate $\sqrt{|x|}$, right? As another hint, if the case $x<0$ confuses you, you might use the fact that if $f(x)$ is an even function (which $\sqrt{|x|}$ is, then$$
H(x) = \int_0^x f(u)~du$$ is an odd function. Draw a graph using that and see what it looks like.

√-x = -(-x)^3/2 ??
so i have two solutions then I can solve for y in each case

 

[LCKurtz]

√-x = -(-x)^3/2 ??
so i have two solutions then I can solve for y in each case

That is true for $x < 0$. So you have a two piece formula for $y$. That is not the same thing as two solutions to your DE. So you have a couple of things left to do. The first is to check whether$$
y = \left \{ \begin{array}{l}
\frac 2 3 x^{\frac 3 2},~ x \ge 0\\
-\frac 2 3 (-x)^{\frac 3 2},~ x < 0
\end{array}\right .$$is a solution to your DE, including at $x=0$. If it is, then you have found one solution. But your question was about uniqueness of solutions to the given problem. Is there another solution?

 

[member 587159]

That is true for $x < 0$. So you have a two piece formula for $y$. That is not the same thing as two solutions to your DE. So you have a couple of things left to do. The first is to check whether$$
y = \left \{ \begin{array}{l}
\frac 2 3 x^{\frac 3 2},~ x \ge 0\\
-\frac 2 3 (-x)^{\frac 3 2},~ x < 0
\end{array}\right .$$is a solution to your DE, including at $x=0$. If it is, then you have found one solution. But your question was about uniqueness of solutions to the given problem. Is there another solution?

Uhm, that's not a solution to the DE. That's the solution to the integral you need to solve to find a solution. Also, I must say to the OP that I made a small mistake in post #2.

The last step should be: $\log_e |y| = \int \sqrt{|x|} dx$ with the absolute values around the y.

 

[nysnacc]

So, y(0)=0 is unique, because both solutions lead to it

 

[member 587159]

So, y(0)=0 is unique, because both solutions lead to it

Show your work please.

 

[nysnacc]

for x>0
ln abs y = 2/3x^3/2
y = A e^(2/3x^3/2) where A is any number

for x<=0

ln abs y = -2/3x^3/2
y = A e^(-2/3x^3/2) where A is any number

for y(0) =0
Any of those equation = 0 where A have to be 0..
so solution is 0 in both case

 

[member 587159]

You made a mistake:

Recalculate:

$\int\sqrt{-x}dx$

 

[nysnacc]

isn't it -2/3 x^3/2 ??

 

[member 587159]

isn't it -2/3 x^3/2 ??

No:

$\int\sqrt{-x}dx$
$-x = u \Rightarrow -du = dx$
$= - \int\sqrt{u}du$
$= - \frac{2}{3}u^{\frac{3}{2}} + c$
$= - \frac{2}{3}(-x)^{\frac{3}{2}} + c$

Anyway, I think you see that you will have a function of the form $y = Ae^{\dots}$. Your job is to find a constant $A$ such that $y(0) = 0$

 

[LCKurtz]

Uhm, that's not a solution to the DE. That's the solution to the integral you need to solve to find a solution.

Correct. I mis-spoke there.

So, y(0)=0 is unique, because both solutions lead to it

It makes no sense to say "y(0)=0 is unique". You are being asked about uniqueness of solutions to a given DE with boundary condition. In other words, is there more than one solution to the given initial value problem. It is trivial to see in this problem that $y \equiv 0$ is a solution. The question is whether there is another different solution. I would think this problem arose in a section where you have an existence and uniqueness theorem that is being discussed, and the question is whether or not it applies to this problem. Do you have such a theorem? Does this problem satisfy its hypotheses? So what can you or can't you say about uniqueness?

 

[nysnacc]

No:

$\int\sqrt{-x}dx$
$-x = u \Rightarrow -du = dx$
$= - \int\sqrt{u}du$
$= - \frac{2}{3}u^{\frac{3}{2}} + c$
$= - \frac{2}{3}(-x)^{\frac{3}{2}} + c$

Anyway, I think you see that you will have a function of the form $y = Ae^{\dots}$. Your job is to find a constant $A$ such that $y(0) = 0$

I found A is 0 in both cases

 

[member 587159]

I found A is 0 in both cases

Okay. That is correct. So, you found the solution $y = 0$, the trivial solution it seems. Now answer to @LCKurtz's questions.

 

[nysnacc]

Okay. That is correct. So, you found the solution $y = 0$, the trivial solution it seems. Now answer to @LCKurtz's questions.

cuz the equation is continuous and differentiable, so it is unique, based on Picard’s theorem on existence and uniqueness

 

[LCKurtz]

cuz the equation is continuous and differentiable, so it is unique, based on Picard’s theorem on existence and uniqueness

That is a very glib and unsatisfactory response. You need to actually address the question. What are the hypotheses of Picard's theorem? State them. Then explain why it does or doesn't apply to this problem.