Second order differential equation question

[issacnewton]

Homework Statement

Verify that $y_1=x^3$ and $y_2=|x|^3$ are linearly independent
solutions of the differential equation $x^2y''-4xy'+6y=0$ on the interval
$(-\infty,\infty)$. Show that $W(y_1,y_2)=0$ for every real number
x, where W is the wronskian.

Homework Equations

theorems on differential equations

The Attempt at a Solution

First I need to check that y₁ and y₂ are the solutions of the
given diff. equation. y₁ is easy. To prove that y₂ is the solution,
I divided the whole interval $(-\infty,\infty)$, in three parts $x>0\; ,x=0\;,\;x>0$. And then I showed that the diff. equation is satisfied on all the different
parts. So that means , y₂ is the solution of the diff. equation

Now, to check the linear independence, let's consider the equation
$$c_1 x^3+c_2 |x|^3=0$$

Now here I am stuck. How do I prove that $c_1=c_2=0$ for all values of x in
$(-\infty,\infty)$.

thanks

 

[Dick]

Homework Statement

Verify that $y_1=x^3$ and $y_2=|x|^3$ are linearly independent
solutions of the differential equation $x^2y''-4xy'+6y=0$ on the interval
$(-\infty,\infty)$. Show that $W(y_1,y_2)=0$ for every real number
x, where W is the wronskian.

Homework Equations

theorems on differential equations

The Attempt at a Solution

First I need to check that y₁ and y₂ are the solutions of the
given diff. equation. y₁ is easy. To prove that y₂ is the solution,
I divided the whole interval $(-\infty,\infty)$, in three parts $x>0\; ,x=0\;,\;x>0$. And then I showed that the diff. equation is satisfied on all the different
parts. So that means , y₂ is the solution of the diff. equation

Now, to check the linear independence, let's consider the equation
$$c_1 x^3+c_2 |x|^3=0$$

Now here I am stuck. How do I prove that $c_1=c_2=0$ for all values of x in
$(-\infty,\infty)$.

thanks

Pick a couple of particular values of x. How about x=1 and x=(-1)?

 

[issacnewton]

Ok, if I plug those values of x, I get $c_1+c_2=0$. But this just means that
c₁ can be expressed in terms of c₂...

Edit: Oh mistake.. another equation is $-c_1 +c_2 = 0$ which gives me
$c_1=0\;\;, c_2=0$ . so $y_1\;,y_2$ are linearly independent.

Now for the remaining part, I have to show that $W(y_1,y_2)=0$ for every real
number. So should I split the interval in 3 parts, since$|x|^3$ is not differentiable on $(-\infty,\infty)$.

 

[LCKurtz]

Ok, if I plug those values of x, I get $c_1+c_2=0$. But this just means that
c₁ can be expressed in terms of c₂...

Edit: Oh mistake.. another equation is $-c_1 +c_2 = 0$ which gives me
$c_1=0\;\;, c_2=0$ . so $y_1\;,y_2$ are linearly independent.

Now for the remaining part, I have to show that $W(y_1,y_2)=0$ for every real
number. So should I split the interval in 3 parts, since$|x|^3$ is not differentiable on $(-\infty,\infty)$.

It's true that $|x|$ isn't differentiable at $x=0$. But you are asserting $|x|^3$ isn't. Are you sure about that? Have you graphed it? Have you checked its right and left hand derivatives?

 

[issacnewton]

Ok I see that $|x|^3$ is differentiable on $(-\infty,\infty)$. But $y_2 '$ is a piecewise function now. So can wronskian be evaluated on different parts of the
interval $(-\infty,\infty)$ ? The examples I have seen with the wronskian don't involve piecewise functions.

 

[Dick]

Ok I see that $|x|^3$ is differentiable on $(-\infty,\infty)$. But $y_2 '$ is a piecewise function now. So can wronskian be evaluated on different parts of the
interval $(-\infty,\infty)$ ? The examples I have seen with the wronskian don't involve piecewise functions.

Just evaluate the wronskian on each piece. Being piecewise defined doesn't really change things.

 

[issacnewton]

Ok that's what I thought. So its valid to evaluate wronskian on different parts of the interval. Can you give the link to such examples... my book on diff. equations (dennis g zill) doesn't have any ...

 

[Dick]

Ok that's what I thought. So its valid to evaluate wronskian on different parts of the interval. Can you give the link to such examples... my book on diff. equations (dennis g zill) doesn't have any ...

I'm not sure why you'd need one. Just try it. If x>=0, y2=x^3 and if x<=0, y2=(-x^3). Just split it up like that.

 

[Bohrok]

By writing |x| as √(x²) and using the chain rule, you can show that $\frac{d}{dx}|x| = \frac{|x|}{x} = \frac{x}{|x|}$ without having to split it up into different cases. Then you can easily show that the Wronskian of x³ and |x|³, after some rewriting, is equal to 0.

 

[issacnewton]

Bohrok, if we write derivative like that, should not we worry about the case when x=0 . That where things get ugly..

 

[Dick]

Bohrok, if we write derivative like that, should not we worry about the case when x=0 . That where things get ugly..

There's nothing ugly about x=0 in the original problem. I don't think Bohrok's hint really simplifies it. Just do it directly.

 

[issacnewton]

thanks... Dick..