finding the general solution for a differential equation

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I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.



I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)



I divided $x^2$ to get rid of the coefficient on $y'$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.



Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance.
$$x^2y'+2xy=5y^4$$
enter image description here






differential-equations







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  • Can you include the DEQ in the actual question for various reasons?
    – Moo
    1 hour ago










  • it's done. Hope it helps
    – Subin Park
    1 hour ago














up vote
1
down vote

favorite
1












I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.



I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)



I divided $x^2$ to get rid of the coefficient on $y'$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.



Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance.
$$x^2y'+2xy=5y^4$$
enter image description here






differential-equations







share|cite|improve this question























  • Can you include the DEQ in the actual question for various reasons?
    – Moo
    1 hour ago










  • it's done. Hope it helps
    – Subin Park
    1 hour ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.



I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)



I divided $x^2$ to get rid of the coefficient on $y'$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.



Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance.
$$x^2y'+2xy=5y^4$$
enter image description here






differential-equations







share|cite|improve this question















I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.



I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)



I divided $x^2$ to get rid of the coefficient on $y'$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.



Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance.
$$x^2y'+2xy=5y^4$$
enter image description here






differential-equations




differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Parcly Taxel

39.8k137097




39.8k137097










asked 1 hour ago









Subin Park

808




808











  • Can you include the DEQ in the actual question for various reasons?
    – Moo
    1 hour ago










  • it's done. Hope it helps
    – Subin Park
    1 hour ago
















  • Can you include the DEQ in the actual question for various reasons?
    – Moo
    1 hour ago










  • it's done. Hope it helps
    – Subin Park
    1 hour ago















Can you include the DEQ in the actual question for various reasons?
– Moo
1 hour ago




Can you include the DEQ in the actual question for various reasons?
– Moo
1 hour ago












it's done. Hope it helps
– Subin Park
1 hour ago




it's done. Hope it helps
– Subin Park
1 hour ago










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










$$x^2y'+2xy=5y^4$$
$$x^2fracy'y^4+2xfrac1y^3=5$$
$$x^2frac-13left(fracy'y^3right)'+2xfrac1y^3=5$$
Obviously the change of function $v=frac1y^3$ is better than $v=y^3$.
$$v=frac1y^3quadtextleads toquad -x^2v'+6xv=15$$
which is a linear ODE, easy to solve :
$$v=cx^6+frac157x$$
$$y^3=frac1cx^6+frac157x$$






share|cite|improve this answer




















  • Hi Jean ! I made a stupid typo ! Cheers.
    – Claude Leibovici
    49 mins ago










  • Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
    – JJacquelin
    15 mins ago

















up vote
3
down vote













$$x^2y'+2xy=5y^4$$
It's really hard to read the picture ...and to point out where you made mistakes
$$(x^2y)'=frac 5y^4x^8 x^8$$
$$int frac dx^2yx^8y^4=5int frac dxx^8$$
$$int frac dx^2y(x^2y)^4=- frac 57x^7+K$$
$$ frac 13(x^2y)^3= frac 57x^7+K$$
$$ x= y^3(frac 157+Cx^7)$$
$$implies y^3= frac x (frac 157+Cx^7)$$






share|cite|improve this answer





























    up vote
    3
    down vote













    Hint



    Being almost blind, it is quite difficult to me to read your notes.



    From what I can see, you used $y=v^3$; use instead $y^3=frac 1v$ and you will arrive to something simple since the equation will become
    $$x^2 v'-6 x v+15=0$$






    share|cite|improve this answer






















    • +1 Claude you aint that blind...I didnt see that ...
      – Isham
      1 hour ago










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    $$x^2y'+2xy=5y^4$$
    $$x^2fracy'y^4+2xfrac1y^3=5$$
    $$x^2frac-13left(fracy'y^3right)'+2xfrac1y^3=5$$
    Obviously the change of function $v=frac1y^3$ is better than $v=y^3$.
    $$v=frac1y^3quadtextleads toquad -x^2v'+6xv=15$$
    which is a linear ODE, easy to solve :
    $$v=cx^6+frac157x$$
    $$y^3=frac1cx^6+frac157x$$






    share|cite|improve this answer




















    • Hi Jean ! I made a stupid typo ! Cheers.
      – Claude Leibovici
      49 mins ago










    • Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
      – JJacquelin
      15 mins ago














    up vote
    2
    down vote



    accepted










    $$x^2y'+2xy=5y^4$$
    $$x^2fracy'y^4+2xfrac1y^3=5$$
    $$x^2frac-13left(fracy'y^3right)'+2xfrac1y^3=5$$
    Obviously the change of function $v=frac1y^3$ is better than $v=y^3$.
    $$v=frac1y^3quadtextleads toquad -x^2v'+6xv=15$$
    which is a linear ODE, easy to solve :
    $$v=cx^6+frac157x$$
    $$y^3=frac1cx^6+frac157x$$






    share|cite|improve this answer




















    • Hi Jean ! I made a stupid typo ! Cheers.
      – Claude Leibovici
      49 mins ago










    • Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
      – JJacquelin
      15 mins ago












    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    $$x^2y'+2xy=5y^4$$
    $$x^2fracy'y^4+2xfrac1y^3=5$$
    $$x^2frac-13left(fracy'y^3right)'+2xfrac1y^3=5$$
    Obviously the change of function $v=frac1y^3$ is better than $v=y^3$.
    $$v=frac1y^3quadtextleads toquad -x^2v'+6xv=15$$
    which is a linear ODE, easy to solve :
    $$v=cx^6+frac157x$$
    $$y^3=frac1cx^6+frac157x$$






    share|cite|improve this answer












    $$x^2y'+2xy=5y^4$$
    $$x^2fracy'y^4+2xfrac1y^3=5$$
    $$x^2frac-13left(fracy'y^3right)'+2xfrac1y^3=5$$
    Obviously the change of function $v=frac1y^3$ is better than $v=y^3$.
    $$v=frac1y^3quadtextleads toquad -x^2v'+6xv=15$$
    which is a linear ODE, easy to solve :
    $$v=cx^6+frac157x$$
    $$y^3=frac1cx^6+frac157x$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 58 mins ago









    JJacquelin

    41.5k21650




    41.5k21650











    • Hi Jean ! I made a stupid typo ! Cheers.
      – Claude Leibovici
      49 mins ago










    • Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
      – JJacquelin
      15 mins ago
















    • Hi Jean ! I made a stupid typo ! Cheers.
      – Claude Leibovici
      49 mins ago










    • Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
      – JJacquelin
      15 mins ago















    Hi Jean ! I made a stupid typo ! Cheers.
    – Claude Leibovici
    49 mins ago




    Hi Jean ! I made a stupid typo ! Cheers.
    – Claude Leibovici
    49 mins ago












    Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
    – JJacquelin
    15 mins ago




    Hi Claude ! I didn't see your answer when I was still typing mine. Seeing it now, we agree, +1 for you. Cheers.
    – JJacquelin
    15 mins ago










    up vote
    3
    down vote













    $$x^2y'+2xy=5y^4$$
    It's really hard to read the picture ...and to point out where you made mistakes
    $$(x^2y)'=frac 5y^4x^8 x^8$$
    $$int frac dx^2yx^8y^4=5int frac dxx^8$$
    $$int frac dx^2y(x^2y)^4=- frac 57x^7+K$$
    $$ frac 13(x^2y)^3= frac 57x^7+K$$
    $$ x= y^3(frac 157+Cx^7)$$
    $$implies y^3= frac x (frac 157+Cx^7)$$






    share|cite|improve this answer


























      up vote
      3
      down vote













      $$x^2y'+2xy=5y^4$$
      It's really hard to read the picture ...and to point out where you made mistakes
      $$(x^2y)'=frac 5y^4x^8 x^8$$
      $$int frac dx^2yx^8y^4=5int frac dxx^8$$
      $$int frac dx^2y(x^2y)^4=- frac 57x^7+K$$
      $$ frac 13(x^2y)^3= frac 57x^7+K$$
      $$ x= y^3(frac 157+Cx^7)$$
      $$implies y^3= frac x (frac 157+Cx^7)$$






      share|cite|improve this answer
























        up vote
        3
        down vote










        up vote
        3
        down vote









        $$x^2y'+2xy=5y^4$$
        It's really hard to read the picture ...and to point out where you made mistakes
        $$(x^2y)'=frac 5y^4x^8 x^8$$
        $$int frac dx^2yx^8y^4=5int frac dxx^8$$
        $$int frac dx^2y(x^2y)^4=- frac 57x^7+K$$
        $$ frac 13(x^2y)^3= frac 57x^7+K$$
        $$ x= y^3(frac 157+Cx^7)$$
        $$implies y^3= frac x (frac 157+Cx^7)$$






        share|cite|improve this answer














        $$x^2y'+2xy=5y^4$$
        It's really hard to read the picture ...and to point out where you made mistakes
        $$(x^2y)'=frac 5y^4x^8 x^8$$
        $$int frac dx^2yx^8y^4=5int frac dxx^8$$
        $$int frac dx^2y(x^2y)^4=- frac 57x^7+K$$
        $$ frac 13(x^2y)^3= frac 57x^7+K$$
        $$ x= y^3(frac 157+Cx^7)$$
        $$implies y^3= frac x (frac 157+Cx^7)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        Isham

        12.3k3929




        12.3k3929




















            up vote
            3
            down vote













            Hint



            Being almost blind, it is quite difficult to me to read your notes.



            From what I can see, you used $y=v^3$; use instead $y^3=frac 1v$ and you will arrive to something simple since the equation will become
            $$x^2 v'-6 x v+15=0$$






            share|cite|improve this answer






















            • +1 Claude you aint that blind...I didnt see that ...
              – Isham
              1 hour ago














            up vote
            3
            down vote













            Hint



            Being almost blind, it is quite difficult to me to read your notes.



            From what I can see, you used $y=v^3$; use instead $y^3=frac 1v$ and you will arrive to something simple since the equation will become
            $$x^2 v'-6 x v+15=0$$






            share|cite|improve this answer






















            • +1 Claude you aint that blind...I didnt see that ...
              – Isham
              1 hour ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            Hint



            Being almost blind, it is quite difficult to me to read your notes.



            From what I can see, you used $y=v^3$; use instead $y^3=frac 1v$ and you will arrive to something simple since the equation will become
            $$x^2 v'-6 x v+15=0$$






            share|cite|improve this answer














            Hint



            Being almost blind, it is quite difficult to me to read your notes.



            From what I can see, you used $y=v^3$; use instead $y^3=frac 1v$ and you will arrive to something simple since the equation will become
            $$x^2 v'-6 x v+15=0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 50 mins ago

























            answered 1 hour ago









            Claude Leibovici

            115k1155130




            115k1155130











            • +1 Claude you aint that blind...I didnt see that ...
              – Isham
              1 hour ago
















            • +1 Claude you aint that blind...I didnt see that ...
              – Isham
              1 hour ago















            +1 Claude you aint that blind...I didnt see that ...
            – Isham
            1 hour ago




            +1 Claude you aint that blind...I didnt see that ...
            – Isham
            1 hour ago

















             

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