Find the general solution of the Euler homogenous equation












0












$begingroup$


"Find the general solution of the Euler homogenous equation
$frac{dy}{dx}=frac{2y-x}{2x-y}$."



Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following:



$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv$ = $intfrac{1}{x} dx$



This leads me to the result of



$frac{3}{2}lnfrac{(v-1)^3}{(v+1)}=ln(x)+c$



Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an $e^c$ term, and I am unsure whether this is correct or not.



Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you.






integration







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  • 2




    $begingroup$
    Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:14
















0












$begingroup$


"Find the general solution of the Euler homogenous equation
$frac{dy}{dx}=frac{2y-x}{2x-y}$."



Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following:



$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv$ = $intfrac{1}{x} dx$



This leads me to the result of



$frac{3}{2}lnfrac{(v-1)^3}{(v+1)}=ln(x)+c$



Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an $e^c$ term, and I am unsure whether this is correct or not.



Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you.






integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:14














0












0








0





$begingroup$


"Find the general solution of the Euler homogenous equation
$frac{dy}{dx}=frac{2y-x}{2x-y}$."



Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following:



$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv$ = $intfrac{1}{x} dx$



This leads me to the result of



$frac{3}{2}lnfrac{(v-1)^3}{(v+1)}=ln(x)+c$



Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an $e^c$ term, and I am unsure whether this is correct or not.



Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you.






integration







share|cite|improve this question









$endgroup$




"Find the general solution of the Euler homogenous equation
$frac{dy}{dx}=frac{2y-x}{2x-y}$."



Through my working out, after introducing a new variable v, where y=xv, I end up having to solve the following:



$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv$ = $intfrac{1}{x} dx$



This leads me to the result of



$frac{3}{2}lnfrac{(v-1)^3}{(v+1)}=ln(x)+c$



Before I begin to rearrange this, to get in terms of y, could someone tell me if I am going in the correct direction because it looks as though I'm going to end up with an $e^c$ term, and I am unsure whether this is correct or not.



Ideally, could someone provide an answer they get from solving this problem - just so I can see if I am going in the correct direction/know if I am correct once I reach an answer? Thank you.






integration




integration






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asked Dec 7 '18 at 1:31









TaylorTaylor

42




42








  • 2




    $begingroup$
    Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:14














  • 2




    $begingroup$
    Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
    $endgroup$
    – Nosrati
    Dec 7 '18 at 3:14








2




2




$begingroup$
Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
$endgroup$
– Nosrati
Dec 7 '18 at 3:14




$begingroup$
Shouldn't be $$frac{1}{2}lnfrac{(v-1)}{(v+1)^3}=ln(x)+c$$
$endgroup$
– Nosrati
Dec 7 '18 at 3:14










2 Answers
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$begingroup$

Solutions found from



$$intfrac{g'(y)),dy}{g(y)}=intfrac{f'(x)),dy}{f(x)}implies log|g(y)|=log|f(x)|+c$$



are equivalent to



$$g(y)=pm e^c f(x).$$



But as $c$ is arbitrary, this is also expressed by



$$g(y)=cf(x).$$




Notice that in the event that $g(y)=0$ or $f(x)=0$ somewhere, we have a singularity, which cannot be crossed. This means that the solution can be made of several pieces, with different integration constants.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv = intfrac{1}{x} dx$$



    Integrating we get,



    $$frac{1}{2}log{(|v-1|)}-frac{3}{2}log{(|v+1|)} = log{|x|}+log{c}$$



    Simplifying,



    $$frac{1}{2}log{left(left|frac{v-1}{(v+1)^3}right|right)}=log{(|cx|)}$$



    Further simplification will give us
    $$frac{v-1}{(v+1)^3}={(cx)}^2$$



    Here $log$ denotes the logarithmic function to the base $e$.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Solutions found from



      $$intfrac{g'(y)),dy}{g(y)}=intfrac{f'(x)),dy}{f(x)}implies log|g(y)|=log|f(x)|+c$$



      are equivalent to



      $$g(y)=pm e^c f(x).$$



      But as $c$ is arbitrary, this is also expressed by



      $$g(y)=cf(x).$$




      Notice that in the event that $g(y)=0$ or $f(x)=0$ somewhere, we have a singularity, which cannot be crossed. This means that the solution can be made of several pieces, with different integration constants.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Solutions found from



        $$intfrac{g'(y)),dy}{g(y)}=intfrac{f'(x)),dy}{f(x)}implies log|g(y)|=log|f(x)|+c$$



        are equivalent to



        $$g(y)=pm e^c f(x).$$



        But as $c$ is arbitrary, this is also expressed by



        $$g(y)=cf(x).$$




        Notice that in the event that $g(y)=0$ or $f(x)=0$ somewhere, we have a singularity, which cannot be crossed. This means that the solution can be made of several pieces, with different integration constants.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Solutions found from



          $$intfrac{g'(y)),dy}{g(y)}=intfrac{f'(x)),dy}{f(x)}implies log|g(y)|=log|f(x)|+c$$



          are equivalent to



          $$g(y)=pm e^c f(x).$$



          But as $c$ is arbitrary, this is also expressed by



          $$g(y)=cf(x).$$




          Notice that in the event that $g(y)=0$ or $f(x)=0$ somewhere, we have a singularity, which cannot be crossed. This means that the solution can be made of several pieces, with different integration constants.






          share|cite|improve this answer









          $endgroup$



          Solutions found from



          $$intfrac{g'(y)),dy}{g(y)}=intfrac{f'(x)),dy}{f(x)}implies log|g(y)|=log|f(x)|+c$$



          are equivalent to



          $$g(y)=pm e^c f(x).$$



          But as $c$ is arbitrary, this is also expressed by



          $$g(y)=cf(x).$$




          Notice that in the event that $g(y)=0$ or $f(x)=0$ somewhere, we have a singularity, which cannot be crossed. This means that the solution can be made of several pieces, with different integration constants.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 10:58









          Yves DaoustYves Daoust

          125k671223




          125k671223























              0












              $begingroup$

              $$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv = intfrac{1}{x} dx$$



              Integrating we get,



              $$frac{1}{2}log{(|v-1|)}-frac{3}{2}log{(|v+1|)} = log{|x|}+log{c}$$



              Simplifying,



              $$frac{1}{2}log{left(left|frac{v-1}{(v+1)^3}right|right)}=log{(|cx|)}$$



              Further simplification will give us
              $$frac{v-1}{(v+1)^3}={(cx)}^2$$



              Here $log$ denotes the logarithmic function to the base $e$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv = intfrac{1}{x} dx$$



                Integrating we get,



                $$frac{1}{2}log{(|v-1|)}-frac{3}{2}log{(|v+1|)} = log{|x|}+log{c}$$



                Simplifying,



                $$frac{1}{2}log{left(left|frac{v-1}{(v+1)^3}right|right)}=log{(|cx|)}$$



                Further simplification will give us
                $$frac{v-1}{(v+1)^3}={(cx)}^2$$



                Here $log$ denotes the logarithmic function to the base $e$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv = intfrac{1}{x} dx$$



                  Integrating we get,



                  $$frac{1}{2}log{(|v-1|)}-frac{3}{2}log{(|v+1|)} = log{|x|}+log{c}$$



                  Simplifying,



                  $$frac{1}{2}log{left(left|frac{v-1}{(v+1)^3}right|right)}=log{(|cx|)}$$



                  Further simplification will give us
                  $$frac{v-1}{(v+1)^3}={(cx)}^2$$



                  Here $log$ denotes the logarithmic function to the base $e$.






                  share|cite|improve this answer











                  $endgroup$



                  $$frac{1}{2}intfrac{1}{v-1}dv-frac{3}{2}intfrac{1}{v+1}dv = intfrac{1}{x} dx$$



                  Integrating we get,



                  $$frac{1}{2}log{(|v-1|)}-frac{3}{2}log{(|v+1|)} = log{|x|}+log{c}$$



                  Simplifying,



                  $$frac{1}{2}log{left(left|frac{v-1}{(v+1)^3}right|right)}=log{(|cx|)}$$



                  Further simplification will give us
                  $$frac{v-1}{(v+1)^3}={(cx)}^2$$



                  Here $log$ denotes the logarithmic function to the base $e$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 11:05

























                  answered Dec 7 '18 at 10:54









                  Thomas ShelbyThomas Shelby

                  2,655421




                  2,655421






























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