Show that $x_{(n)}=0spacespaceforall spacespacespace ninmathbb{Z}$.












0












$begingroup$



Let $a,binmathbb{R}$ and the following differential equation:



$$x''-2ax'+bx=0.$$



with $x(0)=x(1)=0.$



Show that $x(n)=0spacespaceforall spacespacespace ninmathbb{Z}$.




I tried to make a substitution and guess a solution $x=e^{rt}$ but then I can't prove the statement. Any hints/ideas?






ordinary-differential-equations







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$endgroup$












  • $begingroup$
    Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
    $endgroup$
    – jobe
    Jan 9 at 13:27










  • $begingroup$
    Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:27










  • $begingroup$
    @jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
    $endgroup$
    – C. Cristi
    Jan 9 at 13:31












  • $begingroup$
    I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
    $endgroup$
    – Wojowu
    Jan 9 at 14:01
















0












$begingroup$



Let $a,binmathbb{R}$ and the following differential equation:



$$x''-2ax'+bx=0.$$



with $x(0)=x(1)=0.$



Show that $x(n)=0spacespaceforall spacespacespace ninmathbb{Z}$.




I tried to make a substitution and guess a solution $x=e^{rt}$ but then I can't prove the statement. Any hints/ideas?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
    $endgroup$
    – jobe
    Jan 9 at 13:27










  • $begingroup$
    Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:27










  • $begingroup$
    @jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
    $endgroup$
    – C. Cristi
    Jan 9 at 13:31












  • $begingroup$
    I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
    $endgroup$
    – Wojowu
    Jan 9 at 14:01














0












0








0





$begingroup$



Let $a,binmathbb{R}$ and the following differential equation:



$$x''-2ax'+bx=0.$$



with $x(0)=x(1)=0.$



Show that $x(n)=0spacespaceforall spacespacespace ninmathbb{Z}$.




I tried to make a substitution and guess a solution $x=e^{rt}$ but then I can't prove the statement. Any hints/ideas?






ordinary-differential-equations







share|cite|improve this question











$endgroup$





Let $a,binmathbb{R}$ and the following differential equation:



$$x''-2ax'+bx=0.$$



with $x(0)=x(1)=0.$



Show that $x(n)=0spacespaceforall spacespacespace ninmathbb{Z}$.




I tried to make a substitution and guess a solution $x=e^{rt}$ but then I can't prove the statement. Any hints/ideas?






ordinary-differential-equations




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 9 at 14:01









Wojowu

17.3k22665




17.3k22665










asked Jan 9 at 13:18









C. CristiC. Cristi

1,542218




1,542218












  • $begingroup$
    Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
    $endgroup$
    – jobe
    Jan 9 at 13:27










  • $begingroup$
    Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:27










  • $begingroup$
    @jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
    $endgroup$
    – C. Cristi
    Jan 9 at 13:31












  • $begingroup$
    I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
    $endgroup$
    – Wojowu
    Jan 9 at 14:01


















  • $begingroup$
    Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
    $endgroup$
    – jobe
    Jan 9 at 13:27










  • $begingroup$
    Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:27










  • $begingroup$
    @jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
    $endgroup$
    – C. Cristi
    Jan 9 at 13:31












  • $begingroup$
    I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
    $endgroup$
    – Wojowu
    Jan 9 at 14:01
















$begingroup$
Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
$endgroup$
– jobe
Jan 9 at 13:27




$begingroup$
Is $x_{(n)}$ the $n$-derivative of $x$ or the value of $x$ at $n$?
$endgroup$
– jobe
Jan 9 at 13:27












$begingroup$
Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
$endgroup$
– Shubham Johri
Jan 9 at 13:27




$begingroup$
Why are $(0),(1),(n)$ in subscript? Do they indicate something other than the value of $x$ at $0,1,n$?
$endgroup$
– Shubham Johri
Jan 9 at 13:27












$begingroup$
@jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
$endgroup$
– C. Cristi
Jan 9 at 13:31






$begingroup$
@jobe Umm, no, they are the value of $x$ at $n$, I will edit this to make it more clear.
$endgroup$
– C. Cristi
Jan 9 at 13:31














$begingroup$
I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
$endgroup$
– Wojowu
Jan 9 at 14:01




$begingroup$
I have removed the [second-order-logic] tag, since it's not relevant to second-order differential equations.
$endgroup$
– Wojowu
Jan 9 at 14:01










3 Answers
3






active

oldest

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1












$begingroup$

The characteristic equation is $$lambda^2-2alambda+b=0$$if the distinct real roots $lambda_1$ and $lambda_2$ exist we can write$$x=pe^{lambda_1t}+1e^{lambda_2t}$$by substitution we obtain $$p+q=0\pe^{lambda_1}+qe^{lambda_2}=0$$which yields to $$p=q=0$$and $$x(t)=0quad,quad tin Bbb R$$ hence the desired result.



If $lambda_1=lambda_2=lambda$ then we have $$x=(p+qt)e^{lambda t}$$therefore $$p=0\p+q=0$$from which we again obtain $$x(t)=0quad,quad tin Bbb R$$



If $lambda_1$ and $lambda_2$ are complex conjugate and unreal roots, let $$lambda_1=alpha+beta i\lambda_2=alpha-beta i$$ therefore $$x(t)=e^{alpha t}(psinbeta t+qcosbeta t)$$by imposing initial condition we obtain $$q=0\psin beta=0$$ if $beta =lpi$ for some $lin Bbb Z$ then $$x(t)=pe^{alpha t}sin lpi t$$which shows that $$x(n)=0quad,quad nin Bbb Z$$ if not, then $sin betane 0$ and we have $p=0$ which leads to $x(t)=0$ hence the result.






share|cite|improve this answer











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  • $begingroup$
    Thank you. Right on time feedback!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:18










  • $begingroup$
    Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:45










  • $begingroup$
    It should be $qt$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:02










  • $begingroup$
    Thank you . That was useful feedback....
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 19:48





















1












$begingroup$

You get $r^2-2ar+b=0$. This gives $r=apmsqrt{a^2-b}$, giving the general solution as $$x=begin{cases}c_1e^{t(a+sqrt{a^2-b})}+c_2e^{t(a-sqrt{a^2-b})},&a^2>b\(c_1+c_2t)e^{at},&a^2=b\e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big],&a^2<bend{cases}$$You are given $x(0)=0=x(1)$, which give $x=0$ as the only solution for the first two cases.



For the third case, note that we get $c_1=0=c_2sin(sqrt{b-a^2})$. This gives $c_2=0$ or $sqrt{b-a^2}=npi,ninBbb N$. The possible solutions are $x=ke^{at}sin(n pi t),kinBbb R$. Now, note that $x(m)=ke^{an}sin(mnpi)=0 forall minBbb Z$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
    $endgroup$
    – Henning Makholm
    Jan 9 at 13:44












  • $begingroup$
    Yes, I am adding that part
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:45










  • $begingroup$
    Can you add more details about how you obtained the third case?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:50










  • $begingroup$
    @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 18:53






  • 1




    $begingroup$
    @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:00





















0












$begingroup$

Here is a solution not requiring explicitly solving the equation. General theory tells us that a solution to such an equation is determined by its value and value of its derivative at a point. In particular, if $x'(1)=0$, then $x$ is constant zero function, so assume $x'(1)neq 0$.



Consider $y(t)=frac{x'(0)}{x'(1)}x(t+1)$. Then it's easy to see $y$ satisfies the same equation as $x$. Further, $y(0)=x(1)=0$ and $y'(0)=frac{x'(0)}{x'(1)}x'(1)=x'(0)$, from which it follows $y(t)=x(t)$. We can now say $0=x(1)=y(1)=x(1+1)=x(2)$, i.e. $x(2)=0$. We can now finish with induction (note that you have treat negative numbers separately).






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    3 Answers
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    active

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The characteristic equation is $$lambda^2-2alambda+b=0$$if the distinct real roots $lambda_1$ and $lambda_2$ exist we can write$$x=pe^{lambda_1t}+1e^{lambda_2t}$$by substitution we obtain $$p+q=0\pe^{lambda_1}+qe^{lambda_2}=0$$which yields to $$p=q=0$$and $$x(t)=0quad,quad tin Bbb R$$ hence the desired result.



    If $lambda_1=lambda_2=lambda$ then we have $$x=(p+qt)e^{lambda t}$$therefore $$p=0\p+q=0$$from which we again obtain $$x(t)=0quad,quad tin Bbb R$$



    If $lambda_1$ and $lambda_2$ are complex conjugate and unreal roots, let $$lambda_1=alpha+beta i\lambda_2=alpha-beta i$$ therefore $$x(t)=e^{alpha t}(psinbeta t+qcosbeta t)$$by imposing initial condition we obtain $$q=0\psin beta=0$$ if $beta =lpi$ for some $lin Bbb Z$ then $$x(t)=pe^{alpha t}sin lpi t$$which shows that $$x(n)=0quad,quad nin Bbb Z$$ if not, then $sin betane 0$ and we have $p=0$ which leads to $x(t)=0$ hence the result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you. Right on time feedback!
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 18:18










    • $begingroup$
      Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:45










    • $begingroup$
      It should be $qt$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:02










    • $begingroup$
      Thank you . That was useful feedback....
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 19:48


















    1












    $begingroup$

    The characteristic equation is $$lambda^2-2alambda+b=0$$if the distinct real roots $lambda_1$ and $lambda_2$ exist we can write$$x=pe^{lambda_1t}+1e^{lambda_2t}$$by substitution we obtain $$p+q=0\pe^{lambda_1}+qe^{lambda_2}=0$$which yields to $$p=q=0$$and $$x(t)=0quad,quad tin Bbb R$$ hence the desired result.



    If $lambda_1=lambda_2=lambda$ then we have $$x=(p+qt)e^{lambda t}$$therefore $$p=0\p+q=0$$from which we again obtain $$x(t)=0quad,quad tin Bbb R$$



    If $lambda_1$ and $lambda_2$ are complex conjugate and unreal roots, let $$lambda_1=alpha+beta i\lambda_2=alpha-beta i$$ therefore $$x(t)=e^{alpha t}(psinbeta t+qcosbeta t)$$by imposing initial condition we obtain $$q=0\psin beta=0$$ if $beta =lpi$ for some $lin Bbb Z$ then $$x(t)=pe^{alpha t}sin lpi t$$which shows that $$x(n)=0quad,quad nin Bbb Z$$ if not, then $sin betane 0$ and we have $p=0$ which leads to $x(t)=0$ hence the result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you. Right on time feedback!
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 18:18










    • $begingroup$
      Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:45










    • $begingroup$
      It should be $qt$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:02










    • $begingroup$
      Thank you . That was useful feedback....
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 19:48
















    1












    1








    1





    $begingroup$

    The characteristic equation is $$lambda^2-2alambda+b=0$$if the distinct real roots $lambda_1$ and $lambda_2$ exist we can write$$x=pe^{lambda_1t}+1e^{lambda_2t}$$by substitution we obtain $$p+q=0\pe^{lambda_1}+qe^{lambda_2}=0$$which yields to $$p=q=0$$and $$x(t)=0quad,quad tin Bbb R$$ hence the desired result.



    If $lambda_1=lambda_2=lambda$ then we have $$x=(p+qt)e^{lambda t}$$therefore $$p=0\p+q=0$$from which we again obtain $$x(t)=0quad,quad tin Bbb R$$



    If $lambda_1$ and $lambda_2$ are complex conjugate and unreal roots, let $$lambda_1=alpha+beta i\lambda_2=alpha-beta i$$ therefore $$x(t)=e^{alpha t}(psinbeta t+qcosbeta t)$$by imposing initial condition we obtain $$q=0\psin beta=0$$ if $beta =lpi$ for some $lin Bbb Z$ then $$x(t)=pe^{alpha t}sin lpi t$$which shows that $$x(n)=0quad,quad nin Bbb Z$$ if not, then $sin betane 0$ and we have $p=0$ which leads to $x(t)=0$ hence the result.






    share|cite|improve this answer











    $endgroup$



    The characteristic equation is $$lambda^2-2alambda+b=0$$if the distinct real roots $lambda_1$ and $lambda_2$ exist we can write$$x=pe^{lambda_1t}+1e^{lambda_2t}$$by substitution we obtain $$p+q=0\pe^{lambda_1}+qe^{lambda_2}=0$$which yields to $$p=q=0$$and $$x(t)=0quad,quad tin Bbb R$$ hence the desired result.



    If $lambda_1=lambda_2=lambda$ then we have $$x=(p+qt)e^{lambda t}$$therefore $$p=0\p+q=0$$from which we again obtain $$x(t)=0quad,quad tin Bbb R$$



    If $lambda_1$ and $lambda_2$ are complex conjugate and unreal roots, let $$lambda_1=alpha+beta i\lambda_2=alpha-beta i$$ therefore $$x(t)=e^{alpha t}(psinbeta t+qcosbeta t)$$by imposing initial condition we obtain $$q=0\psin beta=0$$ if $beta =lpi$ for some $lin Bbb Z$ then $$x(t)=pe^{alpha t}sin lpi t$$which shows that $$x(n)=0quad,quad nin Bbb Z$$ if not, then $sin betane 0$ and we have $p=0$ which leads to $x(t)=0$ hence the result.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 9 at 19:46

























    answered Jan 9 at 15:09









    Mostafa AyazMostafa Ayaz

    15.4k3939




    15.4k3939












    • $begingroup$
      Thank you. Right on time feedback!
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 18:18










    • $begingroup$
      Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:45










    • $begingroup$
      It should be $qt$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:02










    • $begingroup$
      Thank you . That was useful feedback....
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 19:48




















    • $begingroup$
      Thank you. Right on time feedback!
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 18:18










    • $begingroup$
      Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:45










    • $begingroup$
      It should be $qt$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:02










    • $begingroup$
      Thank you . That was useful feedback....
      $endgroup$
      – Mostafa Ayaz
      Jan 9 at 19:48


















    $begingroup$
    Thank you. Right on time feedback!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:18




    $begingroup$
    Thank you. Right on time feedback!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:18












    $begingroup$
    Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:45




    $begingroup$
    Why is that $lambda_1=lambda_2$ yields to $x=(p+qx)e^{lambda t}$ why is that $qx$?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:45












    $begingroup$
    It should be $qt$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:02




    $begingroup$
    It should be $qt$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:02












    $begingroup$
    Thank you . That was useful feedback....
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 19:48






    $begingroup$
    Thank you . That was useful feedback....
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 19:48













    1












    $begingroup$

    You get $r^2-2ar+b=0$. This gives $r=apmsqrt{a^2-b}$, giving the general solution as $$x=begin{cases}c_1e^{t(a+sqrt{a^2-b})}+c_2e^{t(a-sqrt{a^2-b})},&a^2>b\(c_1+c_2t)e^{at},&a^2=b\e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big],&a^2<bend{cases}$$You are given $x(0)=0=x(1)$, which give $x=0$ as the only solution for the first two cases.



    For the third case, note that we get $c_1=0=c_2sin(sqrt{b-a^2})$. This gives $c_2=0$ or $sqrt{b-a^2}=npi,ninBbb N$. The possible solutions are $x=ke^{at}sin(n pi t),kinBbb R$. Now, note that $x(m)=ke^{an}sin(mnpi)=0 forall minBbb Z$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
      $endgroup$
      – Henning Makholm
      Jan 9 at 13:44












    • $begingroup$
      Yes, I am adding that part
      $endgroup$
      – Shubham Johri
      Jan 9 at 13:45










    • $begingroup$
      Can you add more details about how you obtained the third case?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:50










    • $begingroup$
      @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
      $endgroup$
      – Shubham Johri
      Jan 9 at 18:53






    • 1




      $begingroup$
      @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:00


















    1












    $begingroup$

    You get $r^2-2ar+b=0$. This gives $r=apmsqrt{a^2-b}$, giving the general solution as $$x=begin{cases}c_1e^{t(a+sqrt{a^2-b})}+c_2e^{t(a-sqrt{a^2-b})},&a^2>b\(c_1+c_2t)e^{at},&a^2=b\e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big],&a^2<bend{cases}$$You are given $x(0)=0=x(1)$, which give $x=0$ as the only solution for the first two cases.



    For the third case, note that we get $c_1=0=c_2sin(sqrt{b-a^2})$. This gives $c_2=0$ or $sqrt{b-a^2}=npi,ninBbb N$. The possible solutions are $x=ke^{at}sin(n pi t),kinBbb R$. Now, note that $x(m)=ke^{an}sin(mnpi)=0 forall minBbb Z$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
      $endgroup$
      – Henning Makholm
      Jan 9 at 13:44












    • $begingroup$
      Yes, I am adding that part
      $endgroup$
      – Shubham Johri
      Jan 9 at 13:45










    • $begingroup$
      Can you add more details about how you obtained the third case?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:50










    • $begingroup$
      @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
      $endgroup$
      – Shubham Johri
      Jan 9 at 18:53






    • 1




      $begingroup$
      @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:00
















    1












    1








    1





    $begingroup$

    You get $r^2-2ar+b=0$. This gives $r=apmsqrt{a^2-b}$, giving the general solution as $$x=begin{cases}c_1e^{t(a+sqrt{a^2-b})}+c_2e^{t(a-sqrt{a^2-b})},&a^2>b\(c_1+c_2t)e^{at},&a^2=b\e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big],&a^2<bend{cases}$$You are given $x(0)=0=x(1)$, which give $x=0$ as the only solution for the first two cases.



    For the third case, note that we get $c_1=0=c_2sin(sqrt{b-a^2})$. This gives $c_2=0$ or $sqrt{b-a^2}=npi,ninBbb N$. The possible solutions are $x=ke^{at}sin(n pi t),kinBbb R$. Now, note that $x(m)=ke^{an}sin(mnpi)=0 forall minBbb Z$.






    share|cite|improve this answer











    $endgroup$



    You get $r^2-2ar+b=0$. This gives $r=apmsqrt{a^2-b}$, giving the general solution as $$x=begin{cases}c_1e^{t(a+sqrt{a^2-b})}+c_2e^{t(a-sqrt{a^2-b})},&a^2>b\(c_1+c_2t)e^{at},&a^2=b\e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big],&a^2<bend{cases}$$You are given $x(0)=0=x(1)$, which give $x=0$ as the only solution for the first two cases.



    For the third case, note that we get $c_1=0=c_2sin(sqrt{b-a^2})$. This gives $c_2=0$ or $sqrt{b-a^2}=npi,ninBbb N$. The possible solutions are $x=ke^{at}sin(n pi t),kinBbb R$. Now, note that $x(m)=ke^{an}sin(mnpi)=0 forall minBbb Z$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 9 at 18:53

























    answered Jan 9 at 13:37









    Shubham JohriShubham Johri

    4,785717




    4,785717












    • $begingroup$
      How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
      $endgroup$
      – Henning Makholm
      Jan 9 at 13:44












    • $begingroup$
      Yes, I am adding that part
      $endgroup$
      – Shubham Johri
      Jan 9 at 13:45










    • $begingroup$
      Can you add more details about how you obtained the third case?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:50










    • $begingroup$
      @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
      $endgroup$
      – Shubham Johri
      Jan 9 at 18:53






    • 1




      $begingroup$
      @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:00




















    • $begingroup$
      How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
      $endgroup$
      – Henning Makholm
      Jan 9 at 13:44












    • $begingroup$
      Yes, I am adding that part
      $endgroup$
      – Shubham Johri
      Jan 9 at 13:45










    • $begingroup$
      Can you add more details about how you obtained the third case?
      $endgroup$
      – C. Cristi
      Jan 9 at 18:50










    • $begingroup$
      @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
      $endgroup$
      – Shubham Johri
      Jan 9 at 18:53






    • 1




      $begingroup$
      @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
      $endgroup$
      – Shubham Johri
      Jan 9 at 19:00


















    $begingroup$
    How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
    $endgroup$
    – Henning Makholm
    Jan 9 at 13:44






    $begingroup$
    How about $x=sin(pi t)$ with $a=0$, $b=pi^2$? (This is not a counterexample to the OP's claim, but doesn't fit your claim for case 2).
    $endgroup$
    – Henning Makholm
    Jan 9 at 13:44














    $begingroup$
    Yes, I am adding that part
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:45




    $begingroup$
    Yes, I am adding that part
    $endgroup$
    – Shubham Johri
    Jan 9 at 13:45












    $begingroup$
    Can you add more details about how you obtained the third case?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:50




    $begingroup$
    Can you add more details about how you obtained the third case?
    $endgroup$
    – C. Cristi
    Jan 9 at 18:50












    $begingroup$
    @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 18:53




    $begingroup$
    @C.Cristi As in how I got $e^{at}big[c_1cos(tsqrt{b-a^2})+c_2sin(tsqrt{b-a^2})big]$ or how I got $c_2sin(sqrt{b-a^2})=0$?
    $endgroup$
    – Shubham Johri
    Jan 9 at 18:53




    1




    1




    $begingroup$
    @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:00






    $begingroup$
    @C.Cristi If $a^2<b, r=apm isqrt{b-a^2}implies x=c_1e^{t(a+isqrt{b-a^2})}+c_2e^{t(a-isqrt{b-a^2})}=e^{at}big[c_1e^{itsqrt{b-a^2}}+c_2e^{-itsqrt{b-a^2}}big]$. Using $e^{ix}=cos(x)+isin(x)$, we get$$x=e^{at}big[(c_1+c_2)cos(tsqrt{b-a^2})+(c_1-c_2)isin(tsqrt{b-a^2})big]=e^{at}big[k_1cos(tsqrt{b-a^2})+k_2sin(tsqrt{b-a^2})big]$$
    $endgroup$
    – Shubham Johri
    Jan 9 at 19:00













    0












    $begingroup$

    Here is a solution not requiring explicitly solving the equation. General theory tells us that a solution to such an equation is determined by its value and value of its derivative at a point. In particular, if $x'(1)=0$, then $x$ is constant zero function, so assume $x'(1)neq 0$.



    Consider $y(t)=frac{x'(0)}{x'(1)}x(t+1)$. Then it's easy to see $y$ satisfies the same equation as $x$. Further, $y(0)=x(1)=0$ and $y'(0)=frac{x'(0)}{x'(1)}x'(1)=x'(0)$, from which it follows $y(t)=x(t)$. We can now say $0=x(1)=y(1)=x(1+1)=x(2)$, i.e. $x(2)=0$. We can now finish with induction (note that you have treat negative numbers separately).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is a solution not requiring explicitly solving the equation. General theory tells us that a solution to such an equation is determined by its value and value of its derivative at a point. In particular, if $x'(1)=0$, then $x$ is constant zero function, so assume $x'(1)neq 0$.



      Consider $y(t)=frac{x'(0)}{x'(1)}x(t+1)$. Then it's easy to see $y$ satisfies the same equation as $x$. Further, $y(0)=x(1)=0$ and $y'(0)=frac{x'(0)}{x'(1)}x'(1)=x'(0)$, from which it follows $y(t)=x(t)$. We can now say $0=x(1)=y(1)=x(1+1)=x(2)$, i.e. $x(2)=0$. We can now finish with induction (note that you have treat negative numbers separately).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is a solution not requiring explicitly solving the equation. General theory tells us that a solution to such an equation is determined by its value and value of its derivative at a point. In particular, if $x'(1)=0$, then $x$ is constant zero function, so assume $x'(1)neq 0$.



        Consider $y(t)=frac{x'(0)}{x'(1)}x(t+1)$. Then it's easy to see $y$ satisfies the same equation as $x$. Further, $y(0)=x(1)=0$ and $y'(0)=frac{x'(0)}{x'(1)}x'(1)=x'(0)$, from which it follows $y(t)=x(t)$. We can now say $0=x(1)=y(1)=x(1+1)=x(2)$, i.e. $x(2)=0$. We can now finish with induction (note that you have treat negative numbers separately).






        share|cite|improve this answer









        $endgroup$



        Here is a solution not requiring explicitly solving the equation. General theory tells us that a solution to such an equation is determined by its value and value of its derivative at a point. In particular, if $x'(1)=0$, then $x$ is constant zero function, so assume $x'(1)neq 0$.



        Consider $y(t)=frac{x'(0)}{x'(1)}x(t+1)$. Then it's easy to see $y$ satisfies the same equation as $x$. Further, $y(0)=x(1)=0$ and $y'(0)=frac{x'(0)}{x'(1)}x'(1)=x'(0)$, from which it follows $y(t)=x(t)$. We can now say $0=x(1)=y(1)=x(1+1)=x(2)$, i.e. $x(2)=0$. We can now finish with induction (note that you have treat negative numbers separately).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 14:09









        WojowuWojowu

        17.3k22665




        17.3k22665






























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