How to find integrating factor for this differential equation












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Solve this differential equation $dx/dt=x^2t^3+xt$



This is not exact.I don't find any way to solve this by integrating factor






ordinary-differential-equations







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  • 1




    $begingroup$
    This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
    $endgroup$
    – Dylan
    Jan 25 at 14:47
















1












$begingroup$


Solve this differential equation $dx/dt=x^2t^3+xt$



This is not exact.I don't find any way to solve this by integrating factor






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
    $endgroup$
    – Dylan
    Jan 25 at 14:47














1












1








1





$begingroup$


Solve this differential equation $dx/dt=x^2t^3+xt$



This is not exact.I don't find any way to solve this by integrating factor






ordinary-differential-equations







share|cite|improve this question









$endgroup$




Solve this differential equation $dx/dt=x^2t^3+xt$



This is not exact.I don't find any way to solve this by integrating factor






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










asked Jan 25 at 14:07









Epsilon DeltaEpsilon Delta

876




876








  • 1




    $begingroup$
    This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
    $endgroup$
    – Dylan
    Jan 25 at 14:47














  • 1




    $begingroup$
    This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
    $endgroup$
    – Dylan
    Jan 25 at 14:47








1




1




$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47




$begingroup$
This is a Bernoulli equation. The standard method is to substitute $z = x^{-1}$
$endgroup$
– Dylan
Jan 25 at 14:47










1 Answer
1






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0












$begingroup$

$$frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.



Let $mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)mu dt-mu dx=0$$
The condition to be an exact differential is :
$$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
$$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
$mu(x,t)=f(x)g(t)$
$$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if
$$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
Then
$$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
$$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
An integrator factor is :
$$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
$$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$






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    0












    $begingroup$

    $$frac{dx}{dt}=x^2t^3+xt$$
    $$(x^2t^3+xt)dt-dx=0$$
    Solving thanks to the integrating factor method.



    Let $mu(x,t)$ an integrating factor.
    $$(x^2t^3+xt)mu dt-mu dx=0$$
    The condition to be an exact differential is :
    $$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
    $$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
    In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
    $mu(x,t)=f(x)g(t)$
    $$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
    $$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
    The separation of variables becomes possible if
    $$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
    Then
    $$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
    $$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
    An integrator factor is :
    $$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
    From this it is easy to solve the ODE. The result is :
    $$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
    $$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$frac{dx}{dt}=x^2t^3+xt$$
      $$(x^2t^3+xt)dt-dx=0$$
      Solving thanks to the integrating factor method.



      Let $mu(x,t)$ an integrating factor.
      $$(x^2t^3+xt)mu dt-mu dx=0$$
      The condition to be an exact differential is :
      $$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
      $$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
      In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
      $mu(x,t)=f(x)g(t)$
      $$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
      $$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
      The separation of variables becomes possible if
      $$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
      Then
      $$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
      $$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
      An integrator factor is :
      $$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
      From this it is easy to solve the ODE. The result is :
      $$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
      $$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$frac{dx}{dt}=x^2t^3+xt$$
        $$(x^2t^3+xt)dt-dx=0$$
        Solving thanks to the integrating factor method.



        Let $mu(x,t)$ an integrating factor.
        $$(x^2t^3+xt)mu dt-mu dx=0$$
        The condition to be an exact differential is :
        $$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
        $$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
        In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
        $mu(x,t)=f(x)g(t)$
        $$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
        $$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
        The separation of variables becomes possible if
        $$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
        Then
        $$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
        $$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
        An integrator factor is :
        $$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
        From this it is easy to solve the ODE. The result is :
        $$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
        $$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$






        share|cite|improve this answer









        $endgroup$



        $$frac{dx}{dt}=x^2t^3+xt$$
        $$(x^2t^3+xt)dt-dx=0$$
        Solving thanks to the integrating factor method.



        Let $mu(x,t)$ an integrating factor.
        $$(x^2t^3+xt)mu dt-mu dx=0$$
        The condition to be an exact differential is :
        $$frac{partial}{partial x}left((x^2t^3+xt)mu right) = frac{partial}{partial t}(-mu)$$
        $$frac{partial mu}{partial t}+(x^2t^3+xt)frac{partial mu}{partial x}+(2xt^3+t)mu=0$$
        In the most simple cases $mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $mu(x,y)$ fails. So, we try a more general form, for example
        $mu(x,t)=f(x)g(t)$
        $$f(x)frac{dg}{dt}+(x^2t^3+xt)g(t)frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
        $$frac{1}{tg(t)}frac{dg}{dt}+(xt^2+1)frac{x}{f(x)}frac{df}{dx}+(2xt^2+1)=0$$
        The separation of variables becomes possible if
        $$xt^2frac{x}{f(x)}frac{df}{dx}+2xt^2=0quadimpliesquad frac{x}{f(x)}frac{df}{dx}=-2quadimpliesquad f(x)=frac{1}{x^2}$$
        Then
        $$frac{1}{tg(t)}frac{dg}{dt}+frac{x}{f(x)}frac{df}{dx}+1=0 quad;quad frac{1}{tg(t)}frac{dg}{dt}-2+1=0$$
        $$frac{1}{g(t)}frac{dg}{dt}=tquadimpliesquad g(t)=e^{t^2/2}$$
        An integrator factor is :
        $$mu(x,t)=frac{e^{t^2/2}}{x^2}$$
        From this it is easy to solve the ODE. The result is :
        $$e^{t^2/2}left(frac{1}{x}+t^2-2 right)=c$$
        $$x(t)=frac{-1}{2-t^2+c:e^{-t^2/2}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 11 at 17:10









        JJacquelinJJacquelin

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