I found on an exercise sheet in physics the integration of the equation: $$2 \frac{dX_0}{d\tau} = - \frac{X_0 (\tau)}{C_L(\tau)} \frac{dC_L}{d\tau} - C_L(\tau)$$ This equation can be thought as the following bivariable problem: $$2 f'(x) = - \frac{f(x)}{g(x)} g'(x) - g(x)$$ I've looked at Wolfram Alpha, and at integro-differential sheets and I didn't find solution methods for solving this. Yet, the physics book seemed to imply that it was easy (maybe uninteresting?).

The solution they found is: $$X_0(\tau) = \left( E - \frac{C_0}{2} \int_{1}^{\eta(1)} \sqrt{\frac{C_L(\tau)}{C_0}} d\tau\right) \sqrt{\frac{C_L(\tau)}{C_0}}$$

Rephrased in our formulation: $$f(x) = \left( E - \frac{g(0)}{2} \int_{1}^{\eta(1)} \sqrt{\frac{g(x)}{g(0)}} d\tau\right) \sqrt{\frac{g(x)}{g(0)}}$$

$E$ is the boundary condition $X=E$ where $\eta = 1$.

One can rephrase the problem with this form: $$2 \left( log(f(x)) \right)'(x) = - \left( log(g(x)) \right)'(x) - \frac{g(x)}{f(x)}$$

I would like to know a method, or general methods to solve that kind of equations or this specific equation.

Thanks in advance.

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    $\begingroup$ Rephrase the problem as $$f' + \frac{1}{2}(\ln g)' f = - \frac{1}{2} g$$ and use the integrating factor $$\exp \left(\frac{1}{2} \int (\ln g)' dx \right) = \sqrt{g}$$ to get $$(\sqrt{g} f)' = - \frac{1}{2} g^{3/2} \implies f = - \frac{1}{2} g^{-1/2} \int g^{3/2} dx + C g^{-1/2}$$ (provided I haven't made any mistakes, it's very late here). $\endgroup$ Jun 9, 2021 at 18:00
  • $\begingroup$ The comment of @mattos explains the derivation. Apart from this, the initial (boundary) condition is strange: $X_0$ is a function of $τ$ but the condition is about another variable (or function?) $η$. Where's this $η$ from? $\endgroup$ Jun 10, 2021 at 1:34
  • $\begingroup$ @Saad Eta is a boundary conditions on volumic mass: the volumic mass over the volumic mass at time 0. This explains why eta is valued at 0 at the initial condition. $\endgroup$ Jun 10, 2021 at 8:20
  • $\begingroup$ @PackSciences So it should be $X_0(η)$ instead of $X_0(τ)$? $\endgroup$ Jun 10, 2021 at 8:22
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    $\begingroup$ @PackSciences It's not dummy since it's in $\dfrac{\mathrm d}{\mathrm dτ}$. That's why I had a question in the first place. And there's also $η(1)$ in your expression, which makes it more confusing. $\endgroup$ Jun 10, 2021 at 8:32

1 Answer 1


If $\;f(x)\not=0,\;$ then $$2gff'+f^2g'+g^2f=0,$$ $$(f^2g)' = -g^2f,\tag1$$ $$-\dfrac{(f^2g)'}{f^4g^2} = \dfrac1{f^3},$$ $$\dfrac1{f^2g} = \int\dfrac1{f^3}\,\text dx,$$ $$\color{green}{\mathbf{g(x)=\dfrac1{f^2(x)\int\dfrac1{f^3(x)}\,\text dx}.\tag2}}$$

In the general case, expression in $RHS(2)$ for each given function $\;f(x)\;$ allows to define corresponding function $\;g(x),\;$ if it exists.

The alternative way.

Let us start from $(1)$ under the conditions $$f(1)=E>0,\quad f(x)>0,\quad g(x)\not=0.\tag3$$ From $(3)$ follows that both f and g does not change the signs.

If $\;\mathbf{g(x)>0},\;$ then

$$\dfrac12\dfrac{(f^2g)'}{\sqrt{f^2g\,}} = -\dfrac12g^{\large\,^3/_2},$$ $$f(x)\,\sqrt{g(x)}=C-\dfrac12\int\limits_1^xg^{\large\,^3/_2}(t)\,\text dt,\quad C=E\,\sqrt{g(1)},$$

$$f(x)=\dfrac1{\sqrt{g(x)}}\left(E\,\sqrt{g(1)}-\dfrac12\int\limits_1^xg^{\large\,^3/_2}(t)\,\text dt\right).\tag{4a}$$

If $\;\mathbf{g(x)<0},\;$ then

$$\dfrac12\dfrac{(-f^2g)'}{\sqrt{-f^2g\,}} = \dfrac12(-g) ^{\large\,^3/_2},$$ $$f(x)\,\sqrt{-g(x)}=C+\dfrac12\int\limits_1^x(-g(t)) ^{\large\,^3/_2}\,\text dt,\quad C=E\,\sqrt{-g(1)},$$

$$f(x)=\dfrac1{\sqrt{-g(x)}} \left(E\, \sqrt{-g(1)}+\dfrac12\int\limits_1^x(-g(t)) ^{\large\,^3/_2}\,\text dt\right).\tag{4b}$$

Finally, $$\color{green}{\mathbf{f(x)=\dfrac1{\sqrt{|g(x)|}}\left(E\,\sqrt{|g(1)|}-\dfrac12 \operatorname{sgn}(g(1)) \int\limits_1^x|g(t)|^{\large\,^3/_2}\,\text dt\right).\tag5}}$$


Let $\;g(x)=e^{-2x},\;$ then $$f(x)=e^x\left(E-\dfrac12\int\limits_1^x e^{-3t}\,\text dt\right) =\dfrac16e^x\left(6E-e^{-3}+e^{-3x}\right),$$ $$2f'(x)=\dfrac13e^x\left(6E-e^{-3}-2e^{-3x}\right),$$ $$f(x)\dfrac{g'(x)}{g(x)}=-\dfrac13e^x\left(6E-e^{-3}+e^{-3x}\right),$$ $$2f'(x)+f(x)\dfrac{g'(x)}{g(x)}+g(x)=0,$$ $$f(1)=E.$$ Testing confirms $(5).$

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    $\begingroup$ Your solution makes sense, but we don't find the physical result. Do you know why? Did you read @mattos ' comment on the main post? $\endgroup$ Jun 11, 2021 at 5:24
  • $\begingroup$ @PackSciences Thank you for the comment, the alternative way is added. $\endgroup$ Jun 11, 2021 at 14:07
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    $\begingroup$ From antepenultimate step to the last step, how did you get C - integral = sqrt(E²g(0) - int)? Also, neither of (1) and (2) agree with the solution proposed by the physics book (which is not fundamentally a bad thing). Could it be that the physics book has a typo, just that I think it would be interesting to say it. $\endgroup$ Jun 11, 2021 at 18:42
  • $\begingroup$ @PackSciences Thank you for this comment. Sadly, my answer was not clear. I've tried to fix and detalize it. $\endgroup$ Jun 11, 2021 at 23:05
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    $\begingroup$ I understand your calculations (all the variables are positive, if that can help you). But, the result (4a) is not exactly the one found in the physics book. In the integral, the integrand $g$ is at the 1/2-th power, while in yours, it is at 3/2. So do you disagree with the physics textbook (it could be a typo, I agree with you, your computation steps are fine)? $\endgroup$ Jun 12, 2021 at 5:40

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