Find the integrating factor and solve












2












$begingroup$


Find an integrating factor and solve
$(2x^2y + x),dy + (xy^2 + y),dx = 0$



I checked if it was exact, which it wasn't.



Then I found



$M/Y$ to be $xy + 1$



$N/Y$ to be $2xy + 1,$ but when I tried to put it in the form of



$n N/X - mM/Y$ I got $-2xy,$ which didn't really tell me much about the value of $n$ and $m.$



So I tried over by multiplying the original equation by $x^ny^m$ and didn't get two linear equations at the end but two equations, which still had $xy$ terms and am stuck now. Any help would be appreciated.






ordinary-differential-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Find an integrating factor and solve
    $(2x^2y + x),dy + (xy^2 + y),dx = 0$



    I checked if it was exact, which it wasn't.



    Then I found



    $M/Y$ to be $xy + 1$



    $N/Y$ to be $2xy + 1,$ but when I tried to put it in the form of



    $n N/X - mM/Y$ I got $-2xy,$ which didn't really tell me much about the value of $n$ and $m.$



    So I tried over by multiplying the original equation by $x^ny^m$ and didn't get two linear equations at the end but two equations, which still had $xy$ terms and am stuck now. Any help would be appreciated.






    ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Find an integrating factor and solve
      $(2x^2y + x),dy + (xy^2 + y),dx = 0$



      I checked if it was exact, which it wasn't.



      Then I found



      $M/Y$ to be $xy + 1$



      $N/Y$ to be $2xy + 1,$ but when I tried to put it in the form of



      $n N/X - mM/Y$ I got $-2xy,$ which didn't really tell me much about the value of $n$ and $m.$



      So I tried over by multiplying the original equation by $x^ny^m$ and didn't get two linear equations at the end but two equations, which still had $xy$ terms and am stuck now. Any help would be appreciated.






      ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      Find an integrating factor and solve
      $(2x^2y + x),dy + (xy^2 + y),dx = 0$



      I checked if it was exact, which it wasn't.



      Then I found



      $M/Y$ to be $xy + 1$



      $N/Y$ to be $2xy + 1,$ but when I tried to put it in the form of



      $n N/X - mM/Y$ I got $-2xy,$ which didn't really tell me much about the value of $n$ and $m.$



      So I tried over by multiplying the original equation by $x^ny^m$ and didn't get two linear equations at the end but two equations, which still had $xy$ terms and am stuck now. Any help would be appreciated.






      ordinary-differential-equations




      ordinary-differential-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 18:55









      Adrian Keister

      5,27371933




      5,27371933










      asked Dec 19 '18 at 18:48









      dragophdragoph

      111




      111






















          1 Answer
          1






          active

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          2












          $begingroup$

          Let's do your $x^ny^m$ and see where we get:
          $$underbrace{left(2x^{n+2}y^{m+1}+x^{n+1}y^mright)}_{partial_x},dy+underbrace{left(x^{n+1}y^{m+2}+x^ny^{m+1}right)}_{partial_y},dx=0 $$
          begin{align*}
          2(n+2)x^{n+1}y^{m+1}+(n+1)x^ny^m&overset{text{set}}{=}(m+2)x^{n+1}y^{m+1}+(m+1)x^ny^m \
          2(n+2)xy+n+1&=(m+2)xy+m+1 \
          2(n+2)xy+n&=(m+2)xy+m.
          end{align*}
          So the terms not multiplying $xy$ will have to be equal, forcing $n=m$. Can we get the coefficients of $xy$ to line up? We'd need
          begin{align*}
          2(n+2)&=n+2\
          2n+4&=n+2\
          n&=-2=m.
          end{align*}
          It works! Evidently, the integrating factor is
          $$mu(x,y)=frac{1}{x^2y^2}. $$
          If you multiply through by this, you'll find the equation is now exact.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
            $endgroup$
            – dragoph
            Dec 19 '18 at 19:26













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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let's do your $x^ny^m$ and see where we get:
          $$underbrace{left(2x^{n+2}y^{m+1}+x^{n+1}y^mright)}_{partial_x},dy+underbrace{left(x^{n+1}y^{m+2}+x^ny^{m+1}right)}_{partial_y},dx=0 $$
          begin{align*}
          2(n+2)x^{n+1}y^{m+1}+(n+1)x^ny^m&overset{text{set}}{=}(m+2)x^{n+1}y^{m+1}+(m+1)x^ny^m \
          2(n+2)xy+n+1&=(m+2)xy+m+1 \
          2(n+2)xy+n&=(m+2)xy+m.
          end{align*}
          So the terms not multiplying $xy$ will have to be equal, forcing $n=m$. Can we get the coefficients of $xy$ to line up? We'd need
          begin{align*}
          2(n+2)&=n+2\
          2n+4&=n+2\
          n&=-2=m.
          end{align*}
          It works! Evidently, the integrating factor is
          $$mu(x,y)=frac{1}{x^2y^2}. $$
          If you multiply through by this, you'll find the equation is now exact.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
            $endgroup$
            – dragoph
            Dec 19 '18 at 19:26


















          2












          $begingroup$

          Let's do your $x^ny^m$ and see where we get:
          $$underbrace{left(2x^{n+2}y^{m+1}+x^{n+1}y^mright)}_{partial_x},dy+underbrace{left(x^{n+1}y^{m+2}+x^ny^{m+1}right)}_{partial_y},dx=0 $$
          begin{align*}
          2(n+2)x^{n+1}y^{m+1}+(n+1)x^ny^m&overset{text{set}}{=}(m+2)x^{n+1}y^{m+1}+(m+1)x^ny^m \
          2(n+2)xy+n+1&=(m+2)xy+m+1 \
          2(n+2)xy+n&=(m+2)xy+m.
          end{align*}
          So the terms not multiplying $xy$ will have to be equal, forcing $n=m$. Can we get the coefficients of $xy$ to line up? We'd need
          begin{align*}
          2(n+2)&=n+2\
          2n+4&=n+2\
          n&=-2=m.
          end{align*}
          It works! Evidently, the integrating factor is
          $$mu(x,y)=frac{1}{x^2y^2}. $$
          If you multiply through by this, you'll find the equation is now exact.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
            $endgroup$
            – dragoph
            Dec 19 '18 at 19:26
















          2












          2








          2





          $begingroup$

          Let's do your $x^ny^m$ and see where we get:
          $$underbrace{left(2x^{n+2}y^{m+1}+x^{n+1}y^mright)}_{partial_x},dy+underbrace{left(x^{n+1}y^{m+2}+x^ny^{m+1}right)}_{partial_y},dx=0 $$
          begin{align*}
          2(n+2)x^{n+1}y^{m+1}+(n+1)x^ny^m&overset{text{set}}{=}(m+2)x^{n+1}y^{m+1}+(m+1)x^ny^m \
          2(n+2)xy+n+1&=(m+2)xy+m+1 \
          2(n+2)xy+n&=(m+2)xy+m.
          end{align*}
          So the terms not multiplying $xy$ will have to be equal, forcing $n=m$. Can we get the coefficients of $xy$ to line up? We'd need
          begin{align*}
          2(n+2)&=n+2\
          2n+4&=n+2\
          n&=-2=m.
          end{align*}
          It works! Evidently, the integrating factor is
          $$mu(x,y)=frac{1}{x^2y^2}. $$
          If you multiply through by this, you'll find the equation is now exact.






          share|cite|improve this answer











          $endgroup$



          Let's do your $x^ny^m$ and see where we get:
          $$underbrace{left(2x^{n+2}y^{m+1}+x^{n+1}y^mright)}_{partial_x},dy+underbrace{left(x^{n+1}y^{m+2}+x^ny^{m+1}right)}_{partial_y},dx=0 $$
          begin{align*}
          2(n+2)x^{n+1}y^{m+1}+(n+1)x^ny^m&overset{text{set}}{=}(m+2)x^{n+1}y^{m+1}+(m+1)x^ny^m \
          2(n+2)xy+n+1&=(m+2)xy+m+1 \
          2(n+2)xy+n&=(m+2)xy+m.
          end{align*}
          So the terms not multiplying $xy$ will have to be equal, forcing $n=m$. Can we get the coefficients of $xy$ to line up? We'd need
          begin{align*}
          2(n+2)&=n+2\
          2n+4&=n+2\
          n&=-2=m.
          end{align*}
          It works! Evidently, the integrating factor is
          $$mu(x,y)=frac{1}{x^2y^2}. $$
          If you multiply through by this, you'll find the equation is now exact.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 19:31

























          answered Dec 19 '18 at 19:12









          Adrian KeisterAdrian Keister

          5,27371933




          5,27371933








          • 1




            $begingroup$
            I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
            $endgroup$
            – dragoph
            Dec 19 '18 at 19:26
















          • 1




            $begingroup$
            I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
            $endgroup$
            – dragoph
            Dec 19 '18 at 19:26










          1




          1




          $begingroup$
          I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
          $endgroup$
          – dragoph
          Dec 19 '18 at 19:26






          $begingroup$
          I got the same equation as you of $2(n+2)xy + n = (m+2)xy + m$. I didn't realize you would then set the non multiplying terms and solve. I was expecting I would get something not including xy. Thanks!
          $endgroup$
          – dragoph
          Dec 19 '18 at 19:26




















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