Mixing
Wronskian is not defined
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Suppose a general question: What does it mean that the Wronskian at a certain point is not even defined? take for example two solutions for a second order ODE: $ y_1(x)=frac{1}{x^{3}} $ and $y_2(x)=x^{2}$.
We get that $W=frac{5}{x^{2}}$ which is not defined on $x=0$.
What is the correct explanation?
real-analysis ordinary-differential-equations
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
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add a comment |
$begingroup$
Suppose a general question: What does it mean that the Wronskian at a certain point is not even defined? take for example two solutions for a second order ODE: $ y_1(x)=frac{1}{x^{3}} $ and $y_2(x)=x^{2}$.
We get that $W=frac{5}{x^{2}}$ which is not defined on $x=0$.
What is the correct explanation?
real-analysis ordinary-differential-equations
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
$endgroup$
1
$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
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– DonAntonio
Jan 13 at 20:17
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$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19
add a comment |
$begingroup$
Suppose a general question: What does it mean that the Wronskian at a certain point is not even defined? take for example two solutions for a second order ODE: $ y_1(x)=frac{1}{x^{3}} $ and $y_2(x)=x^{2}$.
We get that $W=frac{5}{x^{2}}$ which is not defined on $x=0$.
What is the correct explanation?
real-analysis ordinary-differential-equations
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
$endgroup$
Suppose a general question: What does it mean that the Wronskian at a certain point is not even defined? take for example two solutions for a second order ODE: $ y_1(x)=frac{1}{x^{3}} $ and $y_2(x)=x^{2}$.
We get that $W=frac{5}{x^{2}}$ which is not defined on $x=0$.
What is the correct explanation?
real-analysis ordinary-differential-equations
real-analysis ordinary-differential-equations
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
asked Jan 13 at 20:10
alexander_yzalexander_yz
165
165
1
$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
$endgroup$
– DonAntonio
Jan 13 at 20:17
$begingroup$
$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19
add a comment |
1
$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
$endgroup$
– DonAntonio
Jan 13 at 20:17
$begingroup$
$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19
1
1
$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
$endgroup$
– DonAntonio
Jan 13 at 20:17
$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
$endgroup$
– DonAntonio
Jan 13 at 20:17
$begingroup$
$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19
$begingroup$
$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19
add a comment |
1 Answer
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The solution to your differential equation is by definition a differentiable function defined on an interval of existence.
In your example the function $y_1(x)=frac{1}{x^3}$ is not defined at $x=0$ so at this point the wronskian is not defined.
We need to stay within the interval of existence.
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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$begingroup$
The solution to your differential equation is by definition a differentiable function defined on an interval of existence.
In your example the function $y_1(x)=frac{1}{x^3}$ is not defined at $x=0$ so at this point the wronskian is not defined.
We need to stay within the interval of existence.
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The solution to your differential equation is by definition a differentiable function defined on an interval of existence.
In your example the function $y_1(x)=frac{1}{x^3}$ is not defined at $x=0$ so at this point the wronskian is not defined.
We need to stay within the interval of existence.
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The solution to your differential equation is by definition a differentiable function defined on an interval of existence.
In your example the function $y_1(x)=frac{1}{x^3}$ is not defined at $x=0$ so at this point the wronskian is not defined.
We need to stay within the interval of existence.
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
The solution to your differential equation is by definition a differentiable function defined on an interval of existence.
In your example the function $y_1(x)=frac{1}{x^3}$ is not defined at $x=0$ so at this point the wronskian is not defined.
We need to stay within the interval of existence.
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 13 at 20:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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$begingroup$
I'm not sure what exactly is your doubt, but I think a good explanation is that thopse solutions to the ODE are not defined in a domain containing zero...which seems pretty obvious.
$endgroup$
– DonAntonio
Jan 13 at 20:17
$begingroup$
$y_1$ is also not defined at $x=0$, so this is not surprising. The Wronskian of linearly independent solutions of a second order linear ODE $y'' + a(x) y' + b(x)=0$ is a constant multiple of $exp(-A(x))$ where $A$ is an antiderivative of $a$. In this case $A(x) = 2 ln(x) + c$, so $a(x) = 2/x$. Again, this is not defined at $x=0$.
$endgroup$
– Robert Israel
Jan 13 at 20:19