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- Mar 5, 2012

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What are the solutions to the following differential equation?

$$\frac{dy}{dx} = \sqrt y \qquad \text{with }y(0)=0$$

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- Mar 5, 2012

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What are the solutions to the following differential equation?

$$\frac{dy}{dx} = \sqrt y \qquad \text{with }y(0)=0$$

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- Jan 26, 2012

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Well, the solution $y=x^{2}/4$ works for $x$ non-negative. However, the condition that $dy/dx \ge 0$, which you can see from the original DE, makes this not work for negative $x$. I don't think you can have a non-trivial solution for negative $x$'s, because then $y$ would have to become negative as you go left, which is not allowed because of the square root. The trivial solution $y=0$ works for all $x$, however. So that's the only infinitely differentiable solution to the DE.

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- Jan 26, 2012

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$$\frac{1}{\sqrt{y}} \text{d} y = \text{d} x$$

$$\int \frac{1}{\sqrt{y}} \text{d} y = \int \text{d} x$$

$$2 \sqrt{y} + c_1 = x + c_2$$

$$2 \sqrt{y} = x + c$$

$$y = \frac{x^2 + 2xc + c^2}{4}$$

Now use the condition $y(0) = 0$, $y \geq 0$ to get $c$ and locate the correct solution.

I probably made a mistake, been a while since I did ODE's.

EDIT: is this a tricky sign question? *runs away*

- Feb 13, 2012

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The example is 'interesting' because the ODE doesn't have a single solution given the condition $\displaystyle y(0)=0$. You can esasily see that, given the ODE written in the form $\displaystyle y^{\ '}= f(y)$ where f(*) is function of the y alone, then if $\displaystyle y_{0}(x)$ is solution, then $\displaystyle y_{0}(x+a)$, a being any real number, is also solution. Given the condition $\displaystyle y(0)=0$ one solution is...

$\displaystyle y_{0}(x)=\begin{cases} \frac{x^{2}}{4} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0 \end{cases}$ (1)

... so that $\displaystyle y_{0}(x-a)$ with a>0 is also solution because it obeys to the same condition. In the following image...

... the functions $\displaystyle y_{0}(x)$ and $\displaystyle y_{1}(x)= y_{0} (x-1)$ are represented...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

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$\displaystyle y^{\ '} = \sqrt{y},\ y(0)=0$ (1)

... is justified by the fact that a first order ODE written in te form...

$\displaystyle y^{\ '}= f(x,y),\ y(x_{0})=y_{0}$ (2)

... admits one and only one solution if and only if in $[x_{0},y_{0}]$ both $f(x,y)$ and $f_{y}(x,y)$ are continous... and that of course isn't verified in (1)...

Kind regards

$\chi$ $\sigma$

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My answer was that the problem statement was wrong... but I was proven wrong!

I ran away too... at first... and went back again (couldn't resist ).

As far as the solutions go, I believe $\chi$ $\sigma$ has said it all.

Originally I also missed Ackbach's observation that the solution does not work for negative x.

And I was puzzling why Bacterius's method did not find all solutions, which is when I made a couple of observations.

When we separate the variables, we bring y into the denominator.

That introduces an extra condition that the original problem does not have: $y \ne 0$.

So afterward we need to consider the possibilities where $y$ can be $0$.

When we square the equation to get rid of $\sqrt y$, we introduce an extra solution.

So afterward we need to discard the solution for $x < 0$.

- Feb 13, 2012

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Odinary Differential Equation Question Help!

The ODE is...

$\displaystyle y^{\ '}= 3\ y^{\frac{2}{3}},\ y(0)= a$ (1)

Also in this case the function ...

$\displaystyle y_{0}(x)=\begin{cases} x^{3} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0 \end{cases}$ (2)

... is solution of (1) and that means that any function $\displaystyle y_{0} (x - c)$ c being a real is also sultion of (1). The consequence is that if $\displaystyle a>0$ the (1) will have one and only one solution and if $\displaystyle a \le 0$ the (1) will have infinite solutions. In the figure...

... is represented the case $\displaystyle a =-1$ with two solution $y_{0}$ and $y_{1}$ ...

Kind regards

$\chi$ $\sigma$

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I'm still confused about this one.A similar and even more 'interesting' example of ODE is proposed in...

Odinary Differential Equation Question Help!

The ODE is...

$\displaystyle y^{\ '}= 3\ y^{\frac{2}{3}},\ y(0)= a$ (1)

Also in this case the function ...

$\displaystyle y_{0}(x)=\begin{cases} x^{3} &\text{if}\ x \ge 0\\ 0 &\text{if}\ x<0 \end{cases}$ (2)

... is solution of (1) and that means that any function $\displaystyle y_{0} (x - c)$ c being a real is also sultion of (1). The consequence is that if $\displaystyle a>0$ the (1) will have one and only one solution and if $\displaystyle a \le 0$ the (1) will have infinite solutions. In the figure...

View attachment 660

... is represented the case $\displaystyle a =-1$ with two solution $y_{0}$ and $y_{1}$ ...

Kind regards

$\chi$ $\sigma$

Seems to me that it all depends on how we define $y^{\frac 2 3}$.

Is it defined for negative y or not?

I'm used to treating it as defined, although most calculators cannot handle it.

Anyway, if $y^{\frac 2 3}$ is not defined for negative y, we cannot get solutions if $a < 0$, since the ODE is not defined then.

If $y^{\frac 2 3}$

Any solution can be constructed from the 3 parts, with either the leftmost or the rightmost part fixed by a value of $a \ne 0$.

And, depending on $a$, the leftmost or rightmost part can also be zero.

If we pick $y=x^3$ for $x < 0$, we get:

$$\begin{aligned}y'&=3x^2 &> 0\\

3y^{\frac 2 3} &= 3(x^3)^{\frac 2 3} = 3x^2 &> 0 \end{aligned}$$

So the ODE is satisfied.

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- Feb 13, 2012

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Sincerly I don't undestand why $\displaystyle y^{\frac{2}{3}}$ cannot be defined for $y<0$... and if it is what is precisely Your objection?... can You give more elements, please?...I'm still confused about this one.

Seems to me that it all depends on how we define $y^{\frac 2 3}$.

Is it defined for negative y or not?

I'm used to treating it as defined, although most calculators cannot handle it.

Anyway, if $y^{\frac 2 3}$ is not defined for negative y, we cannot get solutions if $a < 0$, since the ODE is not defined then.

If $y^{\frac 2 3}$defined for negative y, I believe there is no special restriction.is

Any solution can be constructed from the 3 parts, with either the leftmost or the rightmost part fixed by a value of $a \ne 0$.

And, depending on $a$, the leftmost or rightmost part can also be zero.

If we pick $y=x^3$ for $x < 0$, we get:

$$\begin{aligned}y'&=3x^2 &> 0\\

3y^{\frac 2 3} &= 3(x^3)^{\frac 2 3} = 3x^2 &> 0 \end{aligned}$$

So the ODE is satisfied.

Kind regards

$\chi$ $\sigma$

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- Mar 5, 2012

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Most calculators cannot calculate $(-1)^{\frac{2}{3}}$ as a real number.Sincerly I don't undestand why $\displaystyle y^{\frac{2}{3}}$ cannot be defined for $y<0$... and if it is what is precisely Your objection?... can You give more elements, please?...

Kind regards

$\chi$ $\sigma$

This is because $x^\alpha$ with $\alpha \in \mathbb R_{>0}$ is only defined for $x \ge 0$.

However, the domain of $x^{\frac{2}{3}}$ can be and is usually extended for $x < 0$.

This is only possible for fractions that have an odd number in the denominator in its simplified form.

One may wonder if it is wise to define, because it leads to the failure of a power identity:

$$-1 = (-1)^{\frac 2 3 \cdot \frac 3 2} = ((-1)^{\frac 2 3})^{\frac 3 2} = 1^{\frac 3 2} = 1$$

Anyway, if we do extend the domain to negative numbers, we would have infinitely many solutions for $a>0$ as well.

See for instance this graph which identifies these solutions.