Solving a Cauchy problem, differential equation












2












$begingroup$


I have the following Cauchy problem



begin{cases} y'(x) + frac{1}{x^2-1}y(x) = sqrt{x+1} \ y(0) = 0 end{cases}



I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= frac{1}{x^2-1}$ :



$$int A(x)dx=int frac{1}{x^2-1}dx= frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)+c $$



then I obtain : $$e^{A(x)}=e^{frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)}=Big(frac{|x-1|}{|x+1|}Big)^{frac{1}{2}}=sqrt{frac{|x-1|}{|x+1|} }$$



I have attempted to solve it in this way:



$$ sqrt{frac{|x-1|}{|x+1|} }cdot y'(x) + frac{1}{x^2-1}cdot sqrt{frac{|x-1|}{|x+1|} }y = sqrt{x+1}cdotsqrt{frac{|x-1|}{|x+1|} }$$



$$sqrt{frac{|x-1|}{|x+1|} }*y(x) =int sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



$$y(x) =Big(sqrt{frac{|x-1|}{|x+1|} }Big)^{-1}cdotint sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.






ordinary-differential-equations cauchy-problem







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  • $begingroup$
    Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 15 at 18:15
















2












$begingroup$


I have the following Cauchy problem



begin{cases} y'(x) + frac{1}{x^2-1}y(x) = sqrt{x+1} \ y(0) = 0 end{cases}



I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= frac{1}{x^2-1}$ :



$$int A(x)dx=int frac{1}{x^2-1}dx= frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)+c $$



then I obtain : $$e^{A(x)}=e^{frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)}=Big(frac{|x-1|}{|x+1|}Big)^{frac{1}{2}}=sqrt{frac{|x-1|}{|x+1|} }$$



I have attempted to solve it in this way:



$$ sqrt{frac{|x-1|}{|x+1|} }cdot y'(x) + frac{1}{x^2-1}cdot sqrt{frac{|x-1|}{|x+1|} }y = sqrt{x+1}cdotsqrt{frac{|x-1|}{|x+1|} }$$



$$sqrt{frac{|x-1|}{|x+1|} }*y(x) =int sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



$$y(x) =Big(sqrt{frac{|x-1|}{|x+1|} }Big)^{-1}cdotint sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 15 at 18:15














2












2








2





$begingroup$


I have the following Cauchy problem



begin{cases} y'(x) + frac{1}{x^2-1}y(x) = sqrt{x+1} \ y(0) = 0 end{cases}



I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= frac{1}{x^2-1}$ :



$$int A(x)dx=int frac{1}{x^2-1}dx= frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)+c $$



then I obtain : $$e^{A(x)}=e^{frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)}=Big(frac{|x-1|}{|x+1|}Big)^{frac{1}{2}}=sqrt{frac{|x-1|}{|x+1|} }$$



I have attempted to solve it in this way:



$$ sqrt{frac{|x-1|}{|x+1|} }cdot y'(x) + frac{1}{x^2-1}cdot sqrt{frac{|x-1|}{|x+1|} }y = sqrt{x+1}cdotsqrt{frac{|x-1|}{|x+1|} }$$



$$sqrt{frac{|x-1|}{|x+1|} }*y(x) =int sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



$$y(x) =Big(sqrt{frac{|x-1|}{|x+1|} }Big)^{-1}cdotint sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$




I have the following Cauchy problem



begin{cases} y'(x) + frac{1}{x^2-1}y(x) = sqrt{x+1} \ y(0) = 0 end{cases}



I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= frac{1}{x^2-1}$ :



$$int A(x)dx=int frac{1}{x^2-1}dx= frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)+c $$



then I obtain : $$e^{A(x)}=e^{frac{1}{2} logBig(frac{|x-1|}{|x+1|}Big)}=Big(frac{|x-1|}{|x+1|}Big)^{frac{1}{2}}=sqrt{frac{|x-1|}{|x+1|} }$$



I have attempted to solve it in this way:



$$ sqrt{frac{|x-1|}{|x+1|} }cdot y'(x) + frac{1}{x^2-1}cdot sqrt{frac{|x-1|}{|x+1|} }y = sqrt{x+1}cdotsqrt{frac{|x-1|}{|x+1|} }$$



$$sqrt{frac{|x-1|}{|x+1|} }*y(x) =int sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



$$y(x) =Big(sqrt{frac{|x-1|}{|x+1|} }Big)^{-1}cdotint sqrt{frac{x+1}{|x+1|}}cdotsqrt{|x-1|}dx$$



Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.






ordinary-differential-equations cauchy-problem




ordinary-differential-equations cauchy-problem






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edited Jan 15 at 19:14









Robert Z

101k1070143




101k1070143










asked Jan 15 at 18:07









andrewandrew

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698












  • $begingroup$
    Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 15 at 18:15


















  • $begingroup$
    Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 15 at 18:15
















$begingroup$
Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
$endgroup$
– Dr. Sonnhard Graubner
Jan 15 at 18:15




$begingroup$
Maple says this here $$y left( x right) ={frac { left( x+1 right) {it _C1}}{sqrt {-{x }^{2}+1}}}+2/3, left( x-1 right) sqrt {x+1} $$
$endgroup$
– Dr. Sonnhard Graubner
Jan 15 at 18:15










2 Answers
2






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4












$begingroup$

Note that since you have on the right side $sqrt{x+1}$, we may assume that $xgeq -1$. Moreover the coefficient $frac{1}{x^2-1}$ implies that $xnot=pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
$$sqrt{frac{1-x}{1+x}}cdot y(x) =int sqrt{1-x},dx.$$
Can you take it from here? ... and do not forget the constant of integration!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The condition is $y(0) = 0$
    $endgroup$
    – andrew
    Jan 15 at 18:52












  • $begingroup$
    @andrew OK. Have you tried to go on?
    $endgroup$
    – Robert Z
    Jan 15 at 18:54










  • $begingroup$
    yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
    $endgroup$
    – andrew
    Jan 16 at 18:30












  • $begingroup$
    then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
    $endgroup$
    – andrew
    Jan 16 at 18:30






  • 1




    $begingroup$
    Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
    $endgroup$
    – Robert Z
    Jan 16 at 18:46





















2












$begingroup$

Computing $$mu(x)=e^{intfrac{1}{x^2-1}dx}=frac{sqrt{1-x}}{sqrt{1+x}}$$ then you will get
$$intfrac{d}{dx}left(frac{sqrt{1-x}y(x)}{sqrt{x+1}}right)=intsqrt{1-x}dx$$






share|cite|improve this answer









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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    4












    $begingroup$

    Note that since you have on the right side $sqrt{x+1}$, we may assume that $xgeq -1$. Moreover the coefficient $frac{1}{x^2-1}$ implies that $xnot=pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
    $$sqrt{frac{1-x}{1+x}}cdot y(x) =int sqrt{1-x},dx.$$
    Can you take it from here? ... and do not forget the constant of integration!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The condition is $y(0) = 0$
      $endgroup$
      – andrew
      Jan 15 at 18:52












    • $begingroup$
      @andrew OK. Have you tried to go on?
      $endgroup$
      – Robert Z
      Jan 15 at 18:54










    • $begingroup$
      yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
      $endgroup$
      – andrew
      Jan 16 at 18:30












    • $begingroup$
      then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
      $endgroup$
      – andrew
      Jan 16 at 18:30






    • 1




      $begingroup$
      Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
      $endgroup$
      – Robert Z
      Jan 16 at 18:46


















    4












    $begingroup$

    Note that since you have on the right side $sqrt{x+1}$, we may assume that $xgeq -1$. Moreover the coefficient $frac{1}{x^2-1}$ implies that $xnot=pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
    $$sqrt{frac{1-x}{1+x}}cdot y(x) =int sqrt{1-x},dx.$$
    Can you take it from here? ... and do not forget the constant of integration!






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The condition is $y(0) = 0$
      $endgroup$
      – andrew
      Jan 15 at 18:52












    • $begingroup$
      @andrew OK. Have you tried to go on?
      $endgroup$
      – Robert Z
      Jan 15 at 18:54










    • $begingroup$
      yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
      $endgroup$
      – andrew
      Jan 16 at 18:30












    • $begingroup$
      then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
      $endgroup$
      – andrew
      Jan 16 at 18:30






    • 1




      $begingroup$
      Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
      $endgroup$
      – Robert Z
      Jan 16 at 18:46
















    4












    4








    4





    $begingroup$

    Note that since you have on the right side $sqrt{x+1}$, we may assume that $xgeq -1$. Moreover the coefficient $frac{1}{x^2-1}$ implies that $xnot=pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
    $$sqrt{frac{1-x}{1+x}}cdot y(x) =int sqrt{1-x},dx.$$
    Can you take it from here? ... and do not forget the constant of integration!






    share|cite|improve this answer











    $endgroup$



    Note that since you have on the right side $sqrt{x+1}$, we may assume that $xgeq -1$. Moreover the coefficient $frac{1}{x^2-1}$ implies that $xnot=pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
    $$sqrt{frac{1-x}{1+x}}cdot y(x) =int sqrt{1-x},dx.$$
    Can you take it from here? ... and do not forget the constant of integration!







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 15 at 19:12

























    answered Jan 15 at 18:27









    Robert ZRobert Z

    101k1070143




    101k1070143












    • $begingroup$
      The condition is $y(0) = 0$
      $endgroup$
      – andrew
      Jan 15 at 18:52












    • $begingroup$
      @andrew OK. Have you tried to go on?
      $endgroup$
      – Robert Z
      Jan 15 at 18:54










    • $begingroup$
      yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
      $endgroup$
      – andrew
      Jan 16 at 18:30












    • $begingroup$
      then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
      $endgroup$
      – andrew
      Jan 16 at 18:30






    • 1




      $begingroup$
      Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
      $endgroup$
      – Robert Z
      Jan 16 at 18:46




















    • $begingroup$
      The condition is $y(0) = 0$
      $endgroup$
      – andrew
      Jan 15 at 18:52












    • $begingroup$
      @andrew OK. Have you tried to go on?
      $endgroup$
      – Robert Z
      Jan 15 at 18:54










    • $begingroup$
      yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
      $endgroup$
      – andrew
      Jan 16 at 18:30












    • $begingroup$
      then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
      $endgroup$
      – andrew
      Jan 16 at 18:30






    • 1




      $begingroup$
      Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
      $endgroup$
      – Robert Z
      Jan 16 at 18:46


















    $begingroup$
    The condition is $y(0) = 0$
    $endgroup$
    – andrew
    Jan 15 at 18:52






    $begingroup$
    The condition is $y(0) = 0$
    $endgroup$
    – andrew
    Jan 15 at 18:52














    $begingroup$
    @andrew OK. Have you tried to go on?
    $endgroup$
    – Robert Z
    Jan 15 at 18:54




    $begingroup$
    @andrew OK. Have you tried to go on?
    $endgroup$
    – Robert Z
    Jan 15 at 18:54












    $begingroup$
    yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
    $endgroup$
    – andrew
    Jan 16 at 18:30






    $begingroup$
    yes Robert Z, I obtain that: $$ sqrt{frac{1-x}{x+1} }cdot y'(x) + frac{1}{x^2-1}cdotsqrt{frac{1-x}{x+1} }y = sqrt{x+1}cdotsqrt{frac{1-x}{x+1} }$$ $$ sqrt{frac{1-x}{x+1} }*y(x) = int sqrt{1-x}dx+c$$
    $endgroup$
    – andrew
    Jan 16 at 18:30














    $begingroup$
    then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
    $endgroup$
    – andrew
    Jan 16 at 18:30




    $begingroup$
    then $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ int sqrt{1-x}dx+cBig]$$ $$y(x) =Big(sqrt{frac{1-x}{x+1}}Big)^{-1}*Big[ - frac{2}{3} sqrt{(1-x)^3}+cBig]$$ rembering that $y(0)=0$ well $$ y(0)=-frac{2}{3}+c=0$$ $$ c=frac{2}{3}$$ Then the solution is : $$y(x)=Big(frac{1}{sqrt{frac{1-x}{x+1}}}Big)*[ - frac{2}{3} sqrt{(1-x)^3}+frac{2}{3}Big]$$
    $endgroup$
    – andrew
    Jan 16 at 18:30




    1




    1




    $begingroup$
    Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
    $endgroup$
    – Robert Z
    Jan 16 at 18:46






    $begingroup$
    Looking at your previous questions it seems that you are not aware of the fact that you can can mark one answer as "accepted". Please see math.stackexchange.com/tour
    $endgroup$
    – Robert Z
    Jan 16 at 18:46













    2












    $begingroup$

    Computing $$mu(x)=e^{intfrac{1}{x^2-1}dx}=frac{sqrt{1-x}}{sqrt{1+x}}$$ then you will get
    $$intfrac{d}{dx}left(frac{sqrt{1-x}y(x)}{sqrt{x+1}}right)=intsqrt{1-x}dx$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Computing $$mu(x)=e^{intfrac{1}{x^2-1}dx}=frac{sqrt{1-x}}{sqrt{1+x}}$$ then you will get
      $$intfrac{d}{dx}left(frac{sqrt{1-x}y(x)}{sqrt{x+1}}right)=intsqrt{1-x}dx$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Computing $$mu(x)=e^{intfrac{1}{x^2-1}dx}=frac{sqrt{1-x}}{sqrt{1+x}}$$ then you will get
        $$intfrac{d}{dx}left(frac{sqrt{1-x}y(x)}{sqrt{x+1}}right)=intsqrt{1-x}dx$$






        share|cite|improve this answer









        $endgroup$



        Computing $$mu(x)=e^{intfrac{1}{x^2-1}dx}=frac{sqrt{1-x}}{sqrt{1+x}}$$ then you will get
        $$intfrac{d}{dx}left(frac{sqrt{1-x}y(x)}{sqrt{x+1}}right)=intsqrt{1-x}dx$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 18:20









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        78.4k42867




        78.4k42867






























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