Finding a differential equations given a general solution












0












$begingroup$



Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$




ATTEMPT



Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$



Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be



$$ frac{(n-1)}{x+B}y' - y'' = 0$$



or



$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your equation is wrong, check the degrees of all terms...
    $endgroup$
    – Mindlack
    Jan 5 at 22:55










  • $begingroup$
    I fixed it now.
    $endgroup$
    – James
    Jan 5 at 22:55










  • $begingroup$
    This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
    $endgroup$
    – orion
    Jan 5 at 23:00










  • $begingroup$
    should I expand the binomial ?
    $endgroup$
    – James
    Jan 5 at 23:01










  • $begingroup$
    Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
    $endgroup$
    – Mindlack
    Jan 5 at 23:03
















0












$begingroup$



Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$




ATTEMPT



Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$



Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be



$$ frac{(n-1)}{x+B}y' - y'' = 0$$



or



$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your equation is wrong, check the degrees of all terms...
    $endgroup$
    – Mindlack
    Jan 5 at 22:55










  • $begingroup$
    I fixed it now.
    $endgroup$
    – James
    Jan 5 at 22:55










  • $begingroup$
    This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
    $endgroup$
    – orion
    Jan 5 at 23:00










  • $begingroup$
    should I expand the binomial ?
    $endgroup$
    – James
    Jan 5 at 23:01










  • $begingroup$
    Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
    $endgroup$
    – Mindlack
    Jan 5 at 23:03














0












0








0





$begingroup$



Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$




ATTEMPT



Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$



Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be



$$ frac{(n-1)}{x+B}y' - y'' = 0$$



or



$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?






ordinary-differential-equations







share|cite|improve this question











$endgroup$





Let $A,B in mathbb{R}$. Find a differential equation having the
general solution $$ y(x) = A (x+ B)^n $$




ATTEMPT



Note that $y' = n A (x + B)^{n-1} $. Also, $y'' = n(n-1)A (x+B)^{n-2}$



Note that $(n-1)(x+B)^{-1} y' = y'' $. Thus, a differential equation can be



$$ frac{(n-1)}{x+B}y' - y'' = 0$$



or



$$ (n-1) y' - (x+B) y'' = 0$$
seemts like a trivial problem, but perhaps Im misunderstanding it?
am I correct?






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 5 at 22:55







James

















asked Jan 5 at 22:53









JamesJames

2,638425




2,638425












  • $begingroup$
    Your equation is wrong, check the degrees of all terms...
    $endgroup$
    – Mindlack
    Jan 5 at 22:55










  • $begingroup$
    I fixed it now.
    $endgroup$
    – James
    Jan 5 at 22:55










  • $begingroup$
    This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
    $endgroup$
    – orion
    Jan 5 at 23:00










  • $begingroup$
    should I expand the binomial ?
    $endgroup$
    – James
    Jan 5 at 23:01










  • $begingroup$
    Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
    $endgroup$
    – Mindlack
    Jan 5 at 23:03


















  • $begingroup$
    Your equation is wrong, check the degrees of all terms...
    $endgroup$
    – Mindlack
    Jan 5 at 22:55










  • $begingroup$
    I fixed it now.
    $endgroup$
    – James
    Jan 5 at 22:55










  • $begingroup$
    This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
    $endgroup$
    – orion
    Jan 5 at 23:00










  • $begingroup$
    should I expand the binomial ?
    $endgroup$
    – James
    Jan 5 at 23:01










  • $begingroup$
    Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
    $endgroup$
    – Mindlack
    Jan 5 at 23:03
















$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55




$begingroup$
Your equation is wrong, check the degrees of all terms...
$endgroup$
– Mindlack
Jan 5 at 22:55












$begingroup$
I fixed it now.
$endgroup$
– James
Jan 5 at 22:55




$begingroup$
I fixed it now.
$endgroup$
– James
Jan 5 at 22:55












$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00




$begingroup$
This is a very "soft" question, there is more than one answer. But the problem is that your answer keeps $B$ in it, which should be some sort of integration constant, not a known parameter of the differential equation.
$endgroup$
– orion
Jan 5 at 23:00












$begingroup$
should I expand the binomial ?
$endgroup$
– James
Jan 5 at 23:01




$begingroup$
should I expand the binomial ?
$endgroup$
– James
Jan 5 at 23:01












$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03




$begingroup$
Your differential equation indeed has $A(x+B)^n$ as a solution. It is not, however, the only one, but I do not know whether this is relevant.
$endgroup$
– Mindlack
Jan 5 at 23:03










2 Answers
2






active

oldest

votes


















1












$begingroup$

You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
$$frac{y'}{y}=frac{n}{x+B}$$



This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:



$$x+B=frac{y}{y'}n$$
$$1=nleft(frac{y}{y'}right)'$$
$$1=nfrac{y'^2-yy''}{y'^2}$$
$$nyy''+(1-n)y'^2=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
    $endgroup$
    – James
    Jan 5 at 23:05










  • $begingroup$
    Sure, that works too and gives you the same solution after some manipulation.
    $endgroup$
    – orion
    Jan 5 at 23:07



















1












$begingroup$

Applying Quotient rule for constant , let y= y(x)
$$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
Simplifying
$$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
dfrac{ny}{y^{'}}= {x+B}; $$



Differentiate to eliminate $B$



$$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$



or



$$yy^{''}=y^{'2}(1-1/n) $$



which is the required second order ODE.






share|cite|improve this answer











$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
    $$frac{y'}{y}=frac{n}{x+B}$$



    This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:



    $$x+B=frac{y}{y'}n$$
    $$1=nleft(frac{y}{y'}right)'$$
    $$1=nfrac{y'^2-yy''}{y'^2}$$
    $$nyy''+(1-n)y'^2=0$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
      $endgroup$
      – James
      Jan 5 at 23:05










    • $begingroup$
      Sure, that works too and gives you the same solution after some manipulation.
      $endgroup$
      – orion
      Jan 5 at 23:07
















    1












    $begingroup$

    You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
    $$frac{y'}{y}=frac{n}{x+B}$$



    This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:



    $$x+B=frac{y}{y'}n$$
    $$1=nleft(frac{y}{y'}right)'$$
    $$1=nfrac{y'^2-yy''}{y'^2}$$
    $$nyy''+(1-n)y'^2=0$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
      $endgroup$
      – James
      Jan 5 at 23:05










    • $begingroup$
      Sure, that works too and gives you the same solution after some manipulation.
      $endgroup$
      – orion
      Jan 5 at 23:07














    1












    1








    1





    $begingroup$

    You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
    $$frac{y'}{y}=frac{n}{x+B}$$



    This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:



    $$x+B=frac{y}{y'}n$$
    $$1=nleft(frac{y}{y'}right)'$$
    $$1=nfrac{y'^2-yy''}{y'^2}$$
    $$nyy''+(1-n)y'^2=0$$






    share|cite|improve this answer









    $endgroup$



    You started well, you have $y=A(x+B)^n$ and $y'=An(x+B)^{n-1}$, dividing you get
    $$frac{y'}{y}=frac{n}{x+B}$$



    This is already a differential equation that works for every $A$, but $B$ is still a problem (so far, you need a different differential equation for each $B$). The easiest solution I see is simply expressing it out and differentiating to kill $B$:



    $$x+B=frac{y}{y'}n$$
    $$1=nleft(frac{y}{y'}right)'$$
    $$1=nfrac{y'^2-yy''}{y'^2}$$
    $$nyy''+(1-n)y'^2=0$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 23:04









    orionorion

    13.7k11937




    13.7k11937












    • $begingroup$
      I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
      $endgroup$
      – James
      Jan 5 at 23:05










    • $begingroup$
      Sure, that works too and gives you the same solution after some manipulation.
      $endgroup$
      – orion
      Jan 5 at 23:07


















    • $begingroup$
      I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
      $endgroup$
      – James
      Jan 5 at 23:05










    • $begingroup$
      Sure, that works too and gives you the same solution after some manipulation.
      $endgroup$
      – orion
      Jan 5 at 23:07
















    $begingroup$
    I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
    $endgroup$
    – James
    Jan 5 at 23:05




    $begingroup$
    I was thinking the same but instead I did: write $y^{1/n} = A^{1/n} x + Const $ and differentiating implicitly twice gives one with no A and B
    $endgroup$
    – James
    Jan 5 at 23:05












    $begingroup$
    Sure, that works too and gives you the same solution after some manipulation.
    $endgroup$
    – orion
    Jan 5 at 23:07




    $begingroup$
    Sure, that works too and gives you the same solution after some manipulation.
    $endgroup$
    – orion
    Jan 5 at 23:07











    1












    $begingroup$

    Applying Quotient rule for constant , let y= y(x)
    $$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
    Simplifying
    $$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
    dfrac{ny}{y^{'}}= {x+B}; $$



    Differentiate to eliminate $B$



    $$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$



    or



    $$yy^{''}=y^{'2}(1-1/n) $$



    which is the required second order ODE.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Applying Quotient rule for constant , let y= y(x)
      $$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
      Simplifying
      $$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
      dfrac{ny}{y^{'}}= {x+B}; $$



      Differentiate to eliminate $B$



      $$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$



      or



      $$yy^{''}=y^{'2}(1-1/n) $$



      which is the required second order ODE.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Applying Quotient rule for constant , let y= y(x)
        $$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
        Simplifying
        $$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
        dfrac{ny}{y^{'}}= {x+B}; $$



        Differentiate to eliminate $B$



        $$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$



        or



        $$yy^{''}=y^{'2}(1-1/n) $$



        which is the required second order ODE.






        share|cite|improve this answer











        $endgroup$



        Applying Quotient rule for constant , let y= y(x)
        $$ A= dfrac{y}{(x+B)^n}=dfrac{y^{'}}{n(x+B)^{n-1}} $$
        Simplifying
        $$dfrac{y^{'}}{ny}= dfrac{1}{x+B};quad
        dfrac{ny}{y^{'}}= {x+B}; $$



        Differentiate to eliminate $B$



        $$ndfrac{y^{'2}-yy^{''}}{y^{'2}}=1$$



        or



        $$yy^{''}=y^{'2}(1-1/n) $$



        which is the required second order ODE.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 23:51

























        answered Jan 5 at 23:16









        NarasimhamNarasimham

        21.1k62258




        21.1k62258






























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