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







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












$begingroup$


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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2452534%2ffinding-integrating-factor-for-non-exact-differential-equation-4y-10xdx4x-6%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2452534%2ffinding-integrating-factor-for-non-exact-differential-equation-4y-10xdx4x-6%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Mario Kart Wii

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                                Read more

                                The Binding of Isaac: Rebirth/Afterbirth

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                                Read more

                                Dobbiaco

                                Image
                                Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
                                Read more