Finding integrating factor for non-exact differential equation $(4y-10x)dx+(4x-6x^2y^{-1})dy=0$.












1












$begingroup$


I am given this equation




$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$




where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $mu=x^ny^m$.



I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!






ordinary-differential-equations integrating-factor







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$endgroup$












  • $begingroup$
    Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
    $endgroup$
    – Lord Shark the Unknown
    Oct 1 '17 at 7:51










  • $begingroup$
    After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
    $endgroup$
    – grizzly.bear
    Oct 1 '17 at 8:11










  • $begingroup$
    You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
    $endgroup$
    – Professor Vector
    Oct 1 '17 at 8:19
















1












$begingroup$


I am given this equation




$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$




where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $mu=x^ny^m$.



I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!






ordinary-differential-equations integrating-factor







share|cite|improve this question











$endgroup$












  • $begingroup$
    Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
    $endgroup$
    – Lord Shark the Unknown
    Oct 1 '17 at 7:51










  • $begingroup$
    After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
    $endgroup$
    – grizzly.bear
    Oct 1 '17 at 8:11










  • $begingroup$
    You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
    $endgroup$
    – Professor Vector
    Oct 1 '17 at 8:19














1












1








1





$begingroup$


I am given this equation




$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$




where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $mu=x^ny^m$.



I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!






ordinary-differential-equations integrating-factor







share|cite|improve this question











$endgroup$




I am given this equation




$$(4y-10x)dx+(4x-6x^2y^{-1})dy=0$$




where I must find an integrating factor to turn this into an exact differential. The integrating factor is supposed to be in the form $mu=x^ny^m$.



I have found $M_{y}=4$ and $N_{x}=(4-12xy^{-1})$. It is here where I get stuck. How do I go about finding the integrating factor in the form $mu=x^ny^m$, and what would the $n$ and $m$ end up being? Any help would be greatly appreciated!






ordinary-differential-equations integrating-factor




ordinary-differential-equations integrating-factor






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 1 '17 at 10:09









Nosrati

26.5k62354




26.5k62354










asked Oct 1 '17 at 7:36









grizzly.beargrizzly.bear

16210




16210












  • $begingroup$
    Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
    $endgroup$
    – Lord Shark the Unknown
    Oct 1 '17 at 7:51










  • $begingroup$
    After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
    $endgroup$
    – grizzly.bear
    Oct 1 '17 at 8:11










  • $begingroup$
    You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
    $endgroup$
    – Professor Vector
    Oct 1 '17 at 8:19


















  • $begingroup$
    Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
    $endgroup$
    – Lord Shark the Unknown
    Oct 1 '17 at 7:51










  • $begingroup$
    After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
    $endgroup$
    – grizzly.bear
    Oct 1 '17 at 8:11










  • $begingroup$
    You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
    $endgroup$
    – Professor Vector
    Oct 1 '17 at 8:19
















$begingroup$
Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
$endgroup$
– Lord Shark the Unknown
Oct 1 '17 at 7:51




$begingroup$
Differentiate $x^ny^m(4y-10x)$ with respect to $y$ and $x^ny^m(4x-6x^2/y)$ with respect to $x$. Find, if possible, $m$ and $n$ to ensure these are equal.
$endgroup$
– Lord Shark the Unknown
Oct 1 '17 at 7:51












$begingroup$
After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
$endgroup$
– grizzly.bear
Oct 1 '17 at 8:11




$begingroup$
After differentiating and setting the two terms equal, I have cancelled the term $2y^{m-1}x^{n}$ from both sides. This leaves me with $(-5mx+2my+2y)=(2(n+1)y=3(n+2)x)$. Is there a way to simplify it so that I can find $n$ and $m$?
$endgroup$
– grizzly.bear
Oct 1 '17 at 8:11












$begingroup$
You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
$endgroup$
– Professor Vector
Oct 1 '17 at 8:19




$begingroup$
You can write that as $2(n-m)y+(5m-3n-6)x=0$, and that's possible for all $x,y$ only if $2(n-m)=0$ and $5m-3n-6=0$.
$endgroup$
– Professor Vector
Oct 1 '17 at 8:19










3 Answers
3






active

oldest

votes


















1












$begingroup$

In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12xdfrac{1}{y})$ we have
$$p(z)=dfrac{M_y-N_x}{Ny-Mx}=dfrac{frac{12x}{y}}{4x^2}=dfrac{3}{xy}=dfrac{3}{z}$$
means your integrating factor is of the form $mu(z)=mu(xy)$. So
$$I=e^{int p(z)dz}=z^3=(xy)^3$$
and finally the answer is
$$color{blue}{x^4y^4-2x^5y^3=C}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    write your equation in the form
    $$y'(x)=-frac{y(x)}{x}frac{left(-5+frac{2y(x)}{x}right)}{left(-3+frac{2y(x}{x}right)}$$ and set $$u=frac{y(x)}{x}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form:
      $$frac{dy}{dx}=g(frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble.
      Then $$frac{dy}{dx}=v+xfrac{dv}{dx}$$
      $$v+xfrac{dv}{dx}=g(v)$$
      Which is separable.
      $$frac{dv}{v-g(v)}+frac{1}{x}dx=0$$






      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









        1












        $begingroup$

        In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12xdfrac{1}{y})$ we have
        $$p(z)=dfrac{M_y-N_x}{Ny-Mx}=dfrac{frac{12x}{y}}{4x^2}=dfrac{3}{xy}=dfrac{3}{z}$$
        means your integrating factor is of the form $mu(z)=mu(xy)$. So
        $$I=e^{int p(z)dz}=z^3=(xy)^3$$
        and finally the answer is
        $$color{blue}{x^4y^4-2x^5y^3=C}$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12xdfrac{1}{y})$ we have
          $$p(z)=dfrac{M_y-N_x}{Ny-Mx}=dfrac{frac{12x}{y}}{4x^2}=dfrac{3}{xy}=dfrac{3}{z}$$
          means your integrating factor is of the form $mu(z)=mu(xy)$. So
          $$I=e^{int p(z)dz}=z^3=(xy)^3$$
          and finally the answer is
          $$color{blue}{x^4y^4-2x^5y^3=C}$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12xdfrac{1}{y})$ we have
            $$p(z)=dfrac{M_y-N_x}{Ny-Mx}=dfrac{frac{12x}{y}}{4x^2}=dfrac{3}{xy}=dfrac{3}{z}$$
            means your integrating factor is of the form $mu(z)=mu(xy)$. So
            $$I=e^{int p(z)dz}=z^3=(xy)^3$$
            and finally the answer is
            $$color{blue}{x^4y^4-2x^5y^3=C}$$






            share|cite|improve this answer









            $endgroup$



            In comments you find $x^3y^3$ as integrating factor of your equation. It is nice approach you found and here is another. With $M_{y}=4$ and $N_{x}=(4-12xdfrac{1}{y})$ we have
            $$p(z)=dfrac{M_y-N_x}{Ny-Mx}=dfrac{frac{12x}{y}}{4x^2}=dfrac{3}{xy}=dfrac{3}{z}$$
            means your integrating factor is of the form $mu(z)=mu(xy)$. So
            $$I=e^{int p(z)dz}=z^3=(xy)^3$$
            and finally the answer is
            $$color{blue}{x^4y^4-2x^5y^3=C}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 1 '17 at 10:06









            NosratiNosrati

            26.5k62354




            26.5k62354























                0












                $begingroup$

                write your equation in the form
                $$y'(x)=-frac{y(x)}{x}frac{left(-5+frac{2y(x)}{x}right)}{left(-3+frac{2y(x}{x}right)}$$ and set $$u=frac{y(x)}{x}$$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  write your equation in the form
                  $$y'(x)=-frac{y(x)}{x}frac{left(-5+frac{2y(x)}{x}right)}{left(-3+frac{2y(x}{x}right)}$$ and set $$u=frac{y(x)}{x}$$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    write your equation in the form
                    $$y'(x)=-frac{y(x)}{x}frac{left(-5+frac{2y(x)}{x}right)}{left(-3+frac{2y(x}{x}right)}$$ and set $$u=frac{y(x)}{x}$$






                    share|cite|improve this answer









                    $endgroup$



                    write your equation in the form
                    $$y'(x)=-frac{y(x)}{x}frac{left(-5+frac{2y(x)}{x}right)}{left(-3+frac{2y(x}{x}right)}$$ and set $$u=frac{y(x)}{x}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 1 '17 at 10:09









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    74k42865




                    74k42865























                        0












                        $begingroup$

                        The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form:
                        $$frac{dy}{dx}=g(frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble.
                        Then $$frac{dy}{dx}=v+xfrac{dv}{dx}$$
                        $$v+xfrac{dv}{dx}=g(v)$$
                        Which is separable.
                        $$frac{dv}{v-g(v)}+frac{1}{x}dx=0$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form:
                          $$frac{dy}{dx}=g(frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble.
                          Then $$frac{dy}{dx}=v+xfrac{dv}{dx}$$
                          $$v+xfrac{dv}{dx}=g(v)$$
                          Which is separable.
                          $$frac{dv}{v-g(v)}+frac{1}{x}dx=0$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form:
                            $$frac{dy}{dx}=g(frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble.
                            Then $$frac{dy}{dx}=v+xfrac{dv}{dx}$$
                            $$v+xfrac{dv}{dx}=g(v)$$
                            Which is separable.
                            $$frac{dv}{v-g(v)}+frac{1}{x}dx=0$$






                            share|cite|improve this answer









                            $endgroup$



                            The general form of your DE is $M(x,y)dx+N(x,y)dy=0$, since both $M(x,y)$ and $N(x,y)$ are homogeneous of first order, then your DE can be written in the form:
                            $$frac{dy}{dx}=g(frac{y}{x})$$ Use the transformation $y=vx$ so that your DE becomes separabble.
                            Then $$frac{dy}{dx}=v+xfrac{dv}{dx}$$
                            $$v+xfrac{dv}{dx}=g(v)$$
                            Which is separable.
                            $$frac{dv}{v-g(v)}+frac{1}{x}dx=0$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 9 at 19:32









                            Arsene1412Arsene1412

                            114




                            114






























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