I am stuck on the following Bernoulli equation,

$\frac{dx}{dt}+\frac{1}{4t\ln(t)}$ $x(t)=\frac{t^3}{\ln(t)}$ $x(t)^5$

I have changed it to a first order linear equation

$\frac{dz}{dt}-\frac{1}{t\ln(t)}$ $z=\frac{-4t^3}{\ln(t)}$

Where $z=\frac{1}{x^4}$ and $\frac{dz}{dt}=-4\frac{dx}{dt}\frac{1}{x^5}$

I am now solving using an integrating factor of $\ln(t)$ however I am stuck here. I think I have made a mistake getting to this point as it appears to be unsolvable.

If someone could find my mistake or suggest another way of solving that would be great

  • $\begingroup$ Are you trying an integrating factor of $\ln(t)$, or of $1/\ln(t)$? $\endgroup$ – Teepeemm Apr 15 '18 at 21:23
  • $\begingroup$ I wonder if there is an error. If it had been $\frac{dx}{dt}\mathbf{-}\frac1{4t\ln(t)}x(t)=\frac{t^3}{\ln(t)}x(t)^5$, then the solution would have been expressible in terms of elementary functions. $\endgroup$ – Teepeemm Apr 15 '18 at 21:50

$$x'+\frac{1}{4t\ln(t)}x=\frac{t^3}{\ln(t)}x^5$$ Substitute $z=\frac 1 {x^4}$ $$-z'/4+\frac{1}{4t\ln(t)}z=\frac{t^3}{\ln(t)}$$ $$-z'+\frac{1}{t\ln(t)}z=4\frac{t^3}{\ln(t)}$$ $$(\frac{z}{\ln t})'=-\frac {4{t^3}}{\ln^2 t}$$ $$\frac{z}{\ln t}=-\int \frac {4{t^3}}{\ln^2 t}dt$$ $$z(t)=- \ln t\int \frac {4{t^3}}{\ln^2 t}dt$$ $$\frac 1 {x^4(t)}=- \ln t\int \frac {4{t^3}}{\ln^2 t}dt$$

You can't integrate the last integral with elementary functions

  • 2
    $\begingroup$ Although proceeding from $-z'+\frac1{t\ln(t)}z$ to $(\frac z{\ln t})'$ seems to be the part that was causing OP difficulty. $\endgroup$ – Teepeemm Apr 15 '18 at 21:24
  • $\begingroup$ Maybe @Teepeemm I thought it was the fact that the final integral cant be expressed without Exponential or log integral $\endgroup$ – Isham Apr 15 '18 at 21:31
  • $\begingroup$ @Isham it was the fact the final integral cant be expressed easily which lead to me questioning the use of an integrating factor and my application of the method. Thanks $\endgroup$ – B.L Apr 16 '18 at 11:48
  • $\begingroup$ @B.L That's what I thought since you need special functions to integrate it... It happens sometimes that you cant express an integral with elementary functions $\endgroup$ – Isham Apr 16 '18 at 12:36

You are correct . Now find the integrating factor and continue.

you have $\frac {dz}{dt } -\frac{1}{t\ln(t)}z = \frac{-4t^3}{\ln(t)}$

the integrating factor IF = $\large e^{\int\frac{-1}{t\ln(t)}\,dt}$

there let $\ln(t) =w\implies \frac1{t}\,dt = dw$

IF=$\large e^{\int\frac{-1}{w}\,dw} = \large e^{-\ln(w)}=\large e^{-\ln(\ln(t))}= \large e^{\ln(\frac{1}{\ln(t)})}= \frac{1}{\ln(t)}$

the solution is given by ;

$z\cdot \frac{1}{\ln(t)}=\int\frac{-4t^3}{\ln(t)}\cdot \frac{1}{\ln(t)}\,dt$

Can you continue from here?

EDIT : see @Isham s answer if you want to know about the last part.

  • $\begingroup$ I like the way you leave OP to fight the non-elementary solution appearing in the last step. :) $\endgroup$ – mickep Apr 15 '18 at 18:20
  • $\begingroup$ Thank you :) feels good to be appreciated XD $\endgroup$ – The Integrator Apr 15 '18 at 18:22
  • $\begingroup$ it's $t^3$ and not $t^2$ $\endgroup$ – Isham Apr 15 '18 at 18:27
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    $\begingroup$ oops , thank you . will edit right away $\endgroup$ – The Integrator Apr 15 '18 at 18:27
  • $\begingroup$ +1 @pranavB23 yw..... $\endgroup$ – Isham Apr 15 '18 at 18:41

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