Determine a valid substitution for a differential equation












3












$begingroup$


I am trying to determine whether any of the $v(x)$ substitutions that I'm given are are possible to make the equation first order linear in terms of $v$.



$$y' = frac{y}{x^2} + x^3y^3$$



The given possible substitutions are



$$v(x)=x^3y^3
\v(x)=y^2
\v(x)=y^{-2}
\v(x)=y/x$$



I don't know how to infer the answer by looking at it. So I went with a brute force approach.



Solving for y in the first three yields $y = frac{v^{1/3}}{x}$, $y=sqrt{v}$, and $y=v^{-1/2}$ respectively. I went with the logic that this would not satisfy the linearity of the problem since $v$ in each case is nonlinear.



Is the last one also false because it would make the equation



$$vx = y
\v'x + v = y'
\v'x + v = frac{v}{x} + x^7v^3$$



Thus keeping the equation in a nonlinear form.



This is just my thinking on the matter. I want to make sure it is correct.






ordinary-differential-equations substitution







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I am trying to determine whether any of the $v(x)$ substitutions that I'm given are are possible to make the equation first order linear in terms of $v$.



    $$y' = frac{y}{x^2} + x^3y^3$$



    The given possible substitutions are



    $$v(x)=x^3y^3
    \v(x)=y^2
    \v(x)=y^{-2}
    \v(x)=y/x$$



    I don't know how to infer the answer by looking at it. So I went with a brute force approach.



    Solving for y in the first three yields $y = frac{v^{1/3}}{x}$, $y=sqrt{v}$, and $y=v^{-1/2}$ respectively. I went with the logic that this would not satisfy the linearity of the problem since $v$ in each case is nonlinear.



    Is the last one also false because it would make the equation



    $$vx = y
    \v'x + v = y'
    \v'x + v = frac{v}{x} + x^7v^3$$



    Thus keeping the equation in a nonlinear form.



    This is just my thinking on the matter. I want to make sure it is correct.






    ordinary-differential-equations substitution







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I am trying to determine whether any of the $v(x)$ substitutions that I'm given are are possible to make the equation first order linear in terms of $v$.



      $$y' = frac{y}{x^2} + x^3y^3$$



      The given possible substitutions are



      $$v(x)=x^3y^3
      \v(x)=y^2
      \v(x)=y^{-2}
      \v(x)=y/x$$



      I don't know how to infer the answer by looking at it. So I went with a brute force approach.



      Solving for y in the first three yields $y = frac{v^{1/3}}{x}$, $y=sqrt{v}$, and $y=v^{-1/2}$ respectively. I went with the logic that this would not satisfy the linearity of the problem since $v$ in each case is nonlinear.



      Is the last one also false because it would make the equation



      $$vx = y
      \v'x + v = y'
      \v'x + v = frac{v}{x} + x^7v^3$$



      Thus keeping the equation in a nonlinear form.



      This is just my thinking on the matter. I want to make sure it is correct.






      ordinary-differential-equations substitution







      share|cite|improve this question









      $endgroup$




      I am trying to determine whether any of the $v(x)$ substitutions that I'm given are are possible to make the equation first order linear in terms of $v$.



      $$y' = frac{y}{x^2} + x^3y^3$$



      The given possible substitutions are



      $$v(x)=x^3y^3
      \v(x)=y^2
      \v(x)=y^{-2}
      \v(x)=y/x$$



      I don't know how to infer the answer by looking at it. So I went with a brute force approach.



      Solving for y in the first three yields $y = frac{v^{1/3}}{x}$, $y=sqrt{v}$, and $y=v^{-1/2}$ respectively. I went with the logic that this would not satisfy the linearity of the problem since $v$ in each case is nonlinear.



      Is the last one also false because it would make the equation



      $$vx = y
      \v'x + v = y'
      \v'x + v = frac{v}{x} + x^7v^3$$



      Thus keeping the equation in a nonlinear form.



      This is just my thinking on the matter. I want to make sure it is correct.






      ordinary-differential-equations substitution




      ordinary-differential-equations substitution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      abyssmuabyssmu

      375




      375






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.



          But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as



          $$y' y^{-3} = y^{-2} x^{-2} + x^3$$



          at which point the substitution is pretty quick.



          Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write



          $$yy' = frac{y^2}{x^2} + x^3 y^4 implies frac 1 2 v' = frac 1 {x^2} v + x^3 v^2$$
          and dealing with $v^2$ could well be simpler than dealing with $x^3$.



          Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
            $endgroup$
            – T. Bongers
            4 hours ago










          • $begingroup$
            You were faster than I was...+1
            $endgroup$
            – DonAntonio
            4 hours ago



















          2












          $begingroup$

          Hint: Write your equation in the form $$-frac{2y'(x)}{y(x)^3}+frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=frac{1}{y(x)^2}$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
            $endgroup$
            – T. Bongers
            4 hours ago



















          0












          $begingroup$

          More generally, consider the substitution $v = x^p y^q$ (with $q ne 0$). We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
          If this satisfies a first-order linear equation
          $$ v' = a(x) v + b(x) $$
          that becomes
          $$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
          or
          $$ y' = frac{a(x)x - p}{q x} y + frac{b(x)}{q x^p} y^{1-q} $$
          This can only match with your original equation in the case $q=-2$.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.



            But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as



            $$y' y^{-3} = y^{-2} x^{-2} + x^3$$



            at which point the substitution is pretty quick.



            Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write



            $$yy' = frac{y^2}{x^2} + x^3 y^4 implies frac 1 2 v' = frac 1 {x^2} v + x^3 v^2$$
            and dealing with $v^2$ could well be simpler than dealing with $x^3$.



            Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
              $endgroup$
              – T. Bongers
              4 hours ago










            • $begingroup$
              You were faster than I was...+1
              $endgroup$
              – DonAntonio
              4 hours ago
















            4












            $begingroup$

            The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.



            But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as



            $$y' y^{-3} = y^{-2} x^{-2} + x^3$$



            at which point the substitution is pretty quick.



            Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write



            $$yy' = frac{y^2}{x^2} + x^3 y^4 implies frac 1 2 v' = frac 1 {x^2} v + x^3 v^2$$
            and dealing with $v^2$ could well be simpler than dealing with $x^3$.



            Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
              $endgroup$
              – T. Bongers
              4 hours ago










            • $begingroup$
              You were faster than I was...+1
              $endgroup$
              – DonAntonio
              4 hours ago














            4












            4








            4





            $begingroup$

            The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.



            But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as



            $$y' y^{-3} = y^{-2} x^{-2} + x^3$$



            at which point the substitution is pretty quick.



            Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write



            $$yy' = frac{y^2}{x^2} + x^3 y^4 implies frac 1 2 v' = frac 1 {x^2} v + x^3 v^2$$
            and dealing with $v^2$ could well be simpler than dealing with $x^3$.



            Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.






            share|cite|improve this answer









            $endgroup$



            The problem with $x^3 y^3$ is that its derivative is rather complicated and unnatural to find in the given problem. Using $y/x$ also doesn't work for the same reason.



            But if we take $v = y^{-2}$, its derivative is $-2y^{-3} y'$, which does appear naturally: the stated ODE can be written as



            $$y' y^{-3} = y^{-2} x^{-2} + x^3$$



            at which point the substitution is pretty quick.



            Likewise, $v = y^2$ leads to a simpler (but still non-linear) equation. It's less natural because $2y y'$ doesn't appear obviously... but we could write



            $$yy' = frac{y^2}{x^2} + x^3 y^4 implies frac 1 2 v' = frac 1 {x^2} v + x^3 v^2$$
            and dealing with $v^2$ could well be simpler than dealing with $x^3$.



            Regardless, $v = y^{1 - 3}$ is the standard substitution for a Bernoulli equation like this.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            T. BongersT. Bongers

            23.3k54662




            23.3k54662












            • $begingroup$
              PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
              $endgroup$
              – T. Bongers
              4 hours ago










            • $begingroup$
              You were faster than I was...+1
              $endgroup$
              – DonAntonio
              4 hours ago


















            • $begingroup$
              PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
              $endgroup$
              – T. Bongers
              4 hours ago










            • $begingroup$
              You were faster than I was...+1
              $endgroup$
              – DonAntonio
              4 hours ago
















            $begingroup$
            PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
            $endgroup$
            – T. Bongers
            4 hours ago




            $begingroup$
            PS: I have a slight objection to the phrasing of calling a substitution "false," or saying that some substitutions aren't possible. All these are perfectly valid substitutions, mathematically... the real question is whether they are useful substitutions.
            $endgroup$
            – T. Bongers
            4 hours ago












            $begingroup$
            You were faster than I was...+1
            $endgroup$
            – DonAntonio
            4 hours ago




            $begingroup$
            You were faster than I was...+1
            $endgroup$
            – DonAntonio
            4 hours ago











            2












            $begingroup$

            Hint: Write your equation in the form $$-frac{2y'(x)}{y(x)^3}+frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=frac{1}{y(x)^2}$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
              $endgroup$
              – T. Bongers
              4 hours ago
















            2












            $begingroup$

            Hint: Write your equation in the form $$-frac{2y'(x)}{y(x)^3}+frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=frac{1}{y(x)^2}$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
              $endgroup$
              – T. Bongers
              4 hours ago














            2












            2








            2





            $begingroup$

            Hint: Write your equation in the form $$-frac{2y'(x)}{y(x)^3}+frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=frac{1}{y(x)^2}$$






            share|cite|improve this answer









            $endgroup$



            Hint: Write your equation in the form $$-frac{2y'(x)}{y(x)^3}+frac{2}{x^2y(x)^2}=-2x^3$$ and Substitute $$v(x)=frac{1}{y(x)^2}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            75.4k42865




            75.4k42865








            • 1




              $begingroup$
              I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
              $endgroup$
              – T. Bongers
              4 hours ago














            • 1




              $begingroup$
              I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
              $endgroup$
              – T. Bongers
              4 hours ago








            1




            1




            $begingroup$
            I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
            $endgroup$
            – T. Bongers
            4 hours ago




            $begingroup$
            I don't normally object to things like this, but... this "hint" seems to just be a stripped down version of the answer I wrote well before yours.
            $endgroup$
            – T. Bongers
            4 hours ago











            0












            $begingroup$

            More generally, consider the substitution $v = x^p y^q$ (with $q ne 0$). We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
            If this satisfies a first-order linear equation
            $$ v' = a(x) v + b(x) $$
            that becomes
            $$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
            or
            $$ y' = frac{a(x)x - p}{q x} y + frac{b(x)}{q x^p} y^{1-q} $$
            This can only match with your original equation in the case $q=-2$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              More generally, consider the substitution $v = x^p y^q$ (with $q ne 0$). We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
              If this satisfies a first-order linear equation
              $$ v' = a(x) v + b(x) $$
              that becomes
              $$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
              or
              $$ y' = frac{a(x)x - p}{q x} y + frac{b(x)}{q x^p} y^{1-q} $$
              This can only match with your original equation in the case $q=-2$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                More generally, consider the substitution $v = x^p y^q$ (with $q ne 0$). We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
                If this satisfies a first-order linear equation
                $$ v' = a(x) v + b(x) $$
                that becomes
                $$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
                or
                $$ y' = frac{a(x)x - p}{q x} y + frac{b(x)}{q x^p} y^{1-q} $$
                This can only match with your original equation in the case $q=-2$.






                share|cite|improve this answer









                $endgroup$



                More generally, consider the substitution $v = x^p y^q$ (with $q ne 0$). We have $$v' = p x^{p-1} y^q + q x^p y^{q-1} y'$$
                If this satisfies a first-order linear equation
                $$ v' = a(x) v + b(x) $$
                that becomes
                $$ p x^{p-1} y^q + q x^p y^{q-1} y'= a(x) x^p y^q + b(x) $$
                or
                $$ y' = frac{a(x)x - p}{q x} y + frac{b(x)}{q x^p} y^{1-q} $$
                This can only match with your original equation in the case $q=-2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Robert IsraelRobert Israel

                323k23213467




                323k23213467






























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