Uniqueness of solution of Cauchy-Euler ODE, with conditions












1












$begingroup$



Suppose we have two "similar" non-homogeneous ODE:




  1. $x^2y''-6y=12$

  2. $x^2y''+2xy'-6y=12$


Find a solution that satisfies $y(1)=3$.




I solved the equations as usual and got that the solution families are:




  1. ${y(x) = -2 +frac{c_1}{x^{2}} + c_2x^{3} }$

  2. ${y(x) = -2 +frac{c_1}{x^{3}} + c_2x^{2} }$


Now my question is, how do I determine that the solution that satisfies $y(1)=3$ is unique AND is defined for each ${x in mathbb{R} }$?



I checked the Wronskian for both cases and it turned out that the first case has $W=5$ and the seconds has $W=frac{5}{x^{2}}$, what does it necessarily mean that the Wronskian is not defined at one point? ($x=0$)










share|cite|improve this question











$endgroup$












  • $begingroup$
    "the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
    $endgroup$
    – Did
    Jan 13 at 17:38










  • $begingroup$
    @Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
    $endgroup$
    – alexander_yz
    Jan 13 at 17:43












  • $begingroup$
    Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
    $endgroup$
    – Did
    Jan 13 at 17:45












  • $begingroup$
    @Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
    $endgroup$
    – alexander_yz
    Jan 13 at 17:47










  • $begingroup$
    @alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
    $endgroup$
    – Did
    Jan 13 at 17:53
















1












$begingroup$



Suppose we have two "similar" non-homogeneous ODE:




  1. $x^2y''-6y=12$

  2. $x^2y''+2xy'-6y=12$


Find a solution that satisfies $y(1)=3$.




I solved the equations as usual and got that the solution families are:




  1. ${y(x) = -2 +frac{c_1}{x^{2}} + c_2x^{3} }$

  2. ${y(x) = -2 +frac{c_1}{x^{3}} + c_2x^{2} }$


Now my question is, how do I determine that the solution that satisfies $y(1)=3$ is unique AND is defined for each ${x in mathbb{R} }$?



I checked the Wronskian for both cases and it turned out that the first case has $W=5$ and the seconds has $W=frac{5}{x^{2}}$, what does it necessarily mean that the Wronskian is not defined at one point? ($x=0$)










share|cite|improve this question











$endgroup$












  • $begingroup$
    "the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
    $endgroup$
    – Did
    Jan 13 at 17:38










  • $begingroup$
    @Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
    $endgroup$
    – alexander_yz
    Jan 13 at 17:43












  • $begingroup$
    Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
    $endgroup$
    – Did
    Jan 13 at 17:45












  • $begingroup$
    @Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
    $endgroup$
    – alexander_yz
    Jan 13 at 17:47










  • $begingroup$
    @alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
    $endgroup$
    – Did
    Jan 13 at 17:53














1












1








1





$begingroup$



Suppose we have two "similar" non-homogeneous ODE:




  1. $x^2y''-6y=12$

  2. $x^2y''+2xy'-6y=12$


Find a solution that satisfies $y(1)=3$.




I solved the equations as usual and got that the solution families are:




  1. ${y(x) = -2 +frac{c_1}{x^{2}} + c_2x^{3} }$

  2. ${y(x) = -2 +frac{c_1}{x^{3}} + c_2x^{2} }$


Now my question is, how do I determine that the solution that satisfies $y(1)=3$ is unique AND is defined for each ${x in mathbb{R} }$?



I checked the Wronskian for both cases and it turned out that the first case has $W=5$ and the seconds has $W=frac{5}{x^{2}}$, what does it necessarily mean that the Wronskian is not defined at one point? ($x=0$)










share|cite|improve this question











$endgroup$





Suppose we have two "similar" non-homogeneous ODE:




  1. $x^2y''-6y=12$

  2. $x^2y''+2xy'-6y=12$


Find a solution that satisfies $y(1)=3$.




I solved the equations as usual and got that the solution families are:




  1. ${y(x) = -2 +frac{c_1}{x^{2}} + c_2x^{3} }$

  2. ${y(x) = -2 +frac{c_1}{x^{3}} + c_2x^{2} }$


Now my question is, how do I determine that the solution that satisfies $y(1)=3$ is unique AND is defined for each ${x in mathbb{R} }$?



I checked the Wronskian for both cases and it turned out that the first case has $W=5$ and the seconds has $W=frac{5}{x^{2}}$, what does it necessarily mean that the Wronskian is not defined at one point? ($x=0$)







ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 18:04









Did

248k23225464




248k23225464










asked Jan 13 at 17:28









alexander_yzalexander_yz

165




165












  • $begingroup$
    "the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
    $endgroup$
    – Did
    Jan 13 at 17:38










  • $begingroup$
    @Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
    $endgroup$
    – alexander_yz
    Jan 13 at 17:43












  • $begingroup$
    Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
    $endgroup$
    – Did
    Jan 13 at 17:45












  • $begingroup$
    @Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
    $endgroup$
    – alexander_yz
    Jan 13 at 17:47










  • $begingroup$
    @alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
    $endgroup$
    – Did
    Jan 13 at 17:53


















  • $begingroup$
    "the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
    $endgroup$
    – Did
    Jan 13 at 17:38










  • $begingroup$
    @Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
    $endgroup$
    – alexander_yz
    Jan 13 at 17:43












  • $begingroup$
    Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
    $endgroup$
    – Did
    Jan 13 at 17:45












  • $begingroup$
    @Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
    $endgroup$
    – alexander_yz
    Jan 13 at 17:47










  • $begingroup$
    @alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
    $endgroup$
    – Did
    Jan 13 at 17:53
















$begingroup$
"the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
$endgroup$
– Did
Jan 13 at 17:38




$begingroup$
"the solution that satisfies y(1)=3 is unique" Why do you think it is? By your own work, it is not. "AND is defined for each x∈R" Why do you think it is? Note that this is definitely not what you are asked to show.
$endgroup$
– Did
Jan 13 at 17:38












$begingroup$
@Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
$endgroup$
– alexander_yz
Jan 13 at 17:43






$begingroup$
@Did The real question asks you to find a solution that satisfies $y(1)=3$ and is defined for each ${x in mathbb{R} }$ and to determine whether it is unique or not
$endgroup$
– alexander_yz
Jan 13 at 17:43














$begingroup$
Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
$endgroup$
– Did
Jan 13 at 17:45






$begingroup$
Yeah, the "real question" is not what you wrote in your post. Re the "real question", I fail to understand the problem since you already know all the solutions. Surely you can check that, in both cases, exactly one of these is defined on the whole real line and fits the condition y(1)=3?
$endgroup$
– Did
Jan 13 at 17:45














$begingroup$
@Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
$endgroup$
– alexander_yz
Jan 13 at 17:47




$begingroup$
@Did Well the real question is written in my own words at the last sentence but you thought I was making it up :| I am just wondering whether it is correct to choose $c1=0$ to get rid of the problematic side of not being defined at $x=0$?, Nicely asking :)
$endgroup$
– alexander_yz
Jan 13 at 17:47












$begingroup$
@alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
$endgroup$
– Did
Jan 13 at 17:53




$begingroup$
@alexander_yz Again, you have written yourself everything you need... Look at case 1. Which solutions are defined on the whole real line? Which ones are not? Among those defined on the whole real line, what are/is those/the one such that y(1)=3? Ergo?
$endgroup$
– Did
Jan 13 at 17:53










1 Answer
1






active

oldest

votes


















1












$begingroup$

Sorry for the extended banter. So anyway, here is the correct answer, summarized.



The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook): Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$. Consider the equations:
begin{equation}tag{NH}y''+p(x)y'+q(x)y=f(x)end{equation}
and
begin{equation}tag{H}y''+p(x)y'+q(x)y=0end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$. Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2inmathbb{R}$.



Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,infty)$. Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH). Hence, every solution to (NH) on $(0,infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,infty)$. Similarly, every solution to (NH) on $(-infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since otherwise we would have $lim_{xto 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$. Thus, $y(x)=-2+c_1x^2$ on $(0,infty)$ and $=-2+d_1x^2$ on $(-infty,0)$. Since we require $y(1)=3$, this means $c_1=5$. Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $mathbb{R}$ with $y(1)=3$.



Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=left{begin{array}{ll}-2+d_2x^3,&x<0\-2+5x^3&xgeq 0end{array}right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=left{begin{array}{ll}3d_2x^2,&x<0\15x^2&xgeq 0end{array}right.$$
and hence
$$y''(x)=left{begin{array}{ll}6d_2x,&x<0\30x&xgeq 0end{array}right.$$
so that (1) is satisfied on $mathbb{R}$ with $y(1)=3$.



However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
    $endgroup$
    – alexander_yz
    Jan 13 at 20:00










  • $begingroup$
    @mathcounterexamples.net but we have two different situations
    $endgroup$
    – alexander_yz
    Jan 13 at 20:01











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Sorry for the extended banter. So anyway, here is the correct answer, summarized.



The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook): Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$. Consider the equations:
begin{equation}tag{NH}y''+p(x)y'+q(x)y=f(x)end{equation}
and
begin{equation}tag{H}y''+p(x)y'+q(x)y=0end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$. Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2inmathbb{R}$.



Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,infty)$. Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH). Hence, every solution to (NH) on $(0,infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,infty)$. Similarly, every solution to (NH) on $(-infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since otherwise we would have $lim_{xto 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$. Thus, $y(x)=-2+c_1x^2$ on $(0,infty)$ and $=-2+d_1x^2$ on $(-infty,0)$. Since we require $y(1)=3$, this means $c_1=5$. Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $mathbb{R}$ with $y(1)=3$.



Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=left{begin{array}{ll}-2+d_2x^3,&x<0\-2+5x^3&xgeq 0end{array}right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=left{begin{array}{ll}3d_2x^2,&x<0\15x^2&xgeq 0end{array}right.$$
and hence
$$y''(x)=left{begin{array}{ll}6d_2x,&x<0\30x&xgeq 0end{array}right.$$
so that (1) is satisfied on $mathbb{R}$ with $y(1)=3$.



However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
    $endgroup$
    – alexander_yz
    Jan 13 at 20:00










  • $begingroup$
    @mathcounterexamples.net but we have two different situations
    $endgroup$
    – alexander_yz
    Jan 13 at 20:01
















1












$begingroup$

Sorry for the extended banter. So anyway, here is the correct answer, summarized.



The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook): Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$. Consider the equations:
begin{equation}tag{NH}y''+p(x)y'+q(x)y=f(x)end{equation}
and
begin{equation}tag{H}y''+p(x)y'+q(x)y=0end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$. Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2inmathbb{R}$.



Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,infty)$. Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH). Hence, every solution to (NH) on $(0,infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,infty)$. Similarly, every solution to (NH) on $(-infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since otherwise we would have $lim_{xto 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$. Thus, $y(x)=-2+c_1x^2$ on $(0,infty)$ and $=-2+d_1x^2$ on $(-infty,0)$. Since we require $y(1)=3$, this means $c_1=5$. Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $mathbb{R}$ with $y(1)=3$.



Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=left{begin{array}{ll}-2+d_2x^3,&x<0\-2+5x^3&xgeq 0end{array}right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=left{begin{array}{ll}3d_2x^2,&x<0\15x^2&xgeq 0end{array}right.$$
and hence
$$y''(x)=left{begin{array}{ll}6d_2x,&x<0\30x&xgeq 0end{array}right.$$
so that (1) is satisfied on $mathbb{R}$ with $y(1)=3$.



However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
    $endgroup$
    – alexander_yz
    Jan 13 at 20:00










  • $begingroup$
    @mathcounterexamples.net but we have two different situations
    $endgroup$
    – alexander_yz
    Jan 13 at 20:01














1












1








1





$begingroup$

Sorry for the extended banter. So anyway, here is the correct answer, summarized.



The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook): Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$. Consider the equations:
begin{equation}tag{NH}y''+p(x)y'+q(x)y=f(x)end{equation}
and
begin{equation}tag{H}y''+p(x)y'+q(x)y=0end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$. Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2inmathbb{R}$.



Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,infty)$. Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH). Hence, every solution to (NH) on $(0,infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,infty)$. Similarly, every solution to (NH) on $(-infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since otherwise we would have $lim_{xto 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$. Thus, $y(x)=-2+c_1x^2$ on $(0,infty)$ and $=-2+d_1x^2$ on $(-infty,0)$. Since we require $y(1)=3$, this means $c_1=5$. Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $mathbb{R}$ with $y(1)=3$.



Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=left{begin{array}{ll}-2+d_2x^3,&x<0\-2+5x^3&xgeq 0end{array}right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=left{begin{array}{ll}3d_2x^2,&x<0\15x^2&xgeq 0end{array}right.$$
and hence
$$y''(x)=left{begin{array}{ll}6d_2x,&x<0\30x&xgeq 0end{array}right.$$
so that (1) is satisfied on $mathbb{R}$ with $y(1)=3$.



However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.






share|cite|improve this answer











$endgroup$



Sorry for the extended banter. So anyway, here is the correct answer, summarized.



The theorem that you want to use is the following (it can be found in pretty much every undergraduate DiffEq textbook): Let $I$ be an open interval, and let $p,q,f$ be continuous functions on $I$. Consider the equations:
begin{equation}tag{NH}y''+p(x)y'+q(x)y=f(x)end{equation}
and
begin{equation}tag{H}y''+p(x)y'+q(x)y=0end{equation}
Suppose $y_1$ and $y_2$ are linearly independent solutions to (H) on $I$, and $y_p$ is any solution to (NH) on $I$. Then every solution to (NH) on $I$ has the form
$$y(x)=y_p(x)+c_1y_1(x)+c_2y_2(x)$$
where $c_1,c_2inmathbb{R}$.



Now, rewrite your equation (2) as
$$y''+2x^{-1}y'-6x^{-2}y=12x^{-2}$$
Note that this is (NH) with $p(x)=2x^{-1}$, $q(x)=-6x^{-2}$, and $f(x)=12x^{-2}$, all continuous on $(0,infty)$. Note that $y_1(x)=x^{-3}$ and $y_2(x)=x^2$ are LI solutions to (H) in this case, and that $y_p(x)=-2$ is a solution to (NH). Hence, every solution to (NH) on $(0,infty)$ has the form
$$y_+(x)=-2+c_1x^{-3}+c_2x^2.$$
I'm calling it $y_+$ instead of just $y$ because this is only true for solutions defined on $(0,infty)$. Similarly, every solution to (NH) on $(-infty,0)$ has the form
$$y_-(x)=-2+d_1x^{-3}+d_2x^2.$$
If $y$ is a solution on $mathbb{R}$, then $y$ must agree with $y_-$ and $y_+$ on their domains. Note that $c_1=d_1=0$, since otherwise we would have $lim_{xto 0}y(x)$ either infinite or nonexistent, violating differentiability of $y$. Thus, $y(x)=-2+c_1x^2$ on $(0,infty)$ and $=-2+d_1x^2$ on $(-infty,0)$. Since we require $y(1)=3$, this means $c_1=5$. Note that $y'(x)=10x$ if $x>0$ and $y'(x)=2d_1x$ if $x<0$. Since $y''$ exists, this means $d_1=5$ as well. For continuity we must include $y(0)=-2$ and now we get
$$y(x)=-2+5x^2$$
as the unique solution to (2) on $mathbb{R}$ with $y(1)=3$.



Case (1) is similar, but with one crucial difference. Similar to (2), we find that
$$y(x)=left{begin{array}{ll}-2+d_2x^3,&x<0\-2+5x^3&xgeq 0end{array}right.$$
But now $d_2$ can be any value at all, because we get
$$y'(x)=left{begin{array}{ll}3d_2x^2,&x<0\15x^2&xgeq 0end{array}right.$$
and hence
$$y''(x)=left{begin{array}{ll}6d_2x,&x<0\30x&xgeq 0end{array}right.$$
so that (1) is satisfied on $mathbb{R}$ with $y(1)=3$.



However, note that if we require that $y$ be three times differentiable then $d_2=5$ so the solution is unique in that special case.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 19:19

























answered Jan 13 at 17:35









Ben WBen W

2,276615




2,276615












  • $begingroup$
    @BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
    $endgroup$
    – alexander_yz
    Jan 13 at 20:00










  • $begingroup$
    @mathcounterexamples.net but we have two different situations
    $endgroup$
    – alexander_yz
    Jan 13 at 20:01


















  • $begingroup$
    @BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
    $endgroup$
    – alexander_yz
    Jan 13 at 20:00










  • $begingroup$
    @mathcounterexamples.net but we have two different situations
    $endgroup$
    – alexander_yz
    Jan 13 at 20:01
















$begingroup$
@BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
$endgroup$
– alexander_yz
Jan 13 at 20:00




$begingroup$
@BenW Thank you so much for the thoroughly analysed answer, I can clearly understand the answer and the differences right now!
$endgroup$
– alexander_yz
Jan 13 at 20:00












$begingroup$
@mathcounterexamples.net but we have two different situations
$endgroup$
– alexander_yz
Jan 13 at 20:01




$begingroup$
@mathcounterexamples.net but we have two different situations
$endgroup$
– alexander_yz
Jan 13 at 20:01


















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