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I came across the following type of differential equation while solving a configuration in physics (specifically - viscous fluids) $$\frac {d^2y}{dx^2} + \frac {1}{y} = 1$$ Where $y = f(x)$

The physics problem is not relevant here, I put my question here to ask if there is a nice solution to this. I don't have much formal knowledge of solving a second order D.E. and after some attempts, couldn't get anywhere.

Any help (either to solve this, or to convince me that there is no elementary solution) would be greatly appreciated.

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  • $\begingroup$ This says WA:$$\text{Solve}\left[\int _1^{y(x)}\frac{1}{\sqrt{c_1+2 (K[1]-\log (K[1]))}}dK[1]{}^2=(c_2+x){}^2,y(x)\right]$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 13 '19 at 14:42
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There is an answer, it's just not nice. Multiply both sides by $y'$:

$$y'y'' +\frac{y'}{y} = y' \implies \frac{1}{2}(y')^2 + \log y = y + C_1$$

$$\sqrt{2}x + C_2 = \pm\int \frac{1}{\sqrt{y - \log y +C_1 }}dy$$

for the nonconstant solution and choosing a definite sign for the derivative. We can do this without ambiguity because $y'$ from the equation above can never be $0$ ($\log y < y$) for positive choices of constant $C_1$. For negative choices I would imagine that there are oscillatory solutions.

However there is a constant solution we lost by multiplying both sides by $y'$ in the very beginning. That solution is $y=1$.

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    $\begingroup$ You should have another constant of integration: $$\frac{1}{2}(y')^2+\ln(y)=y+k\,.$$ $\endgroup$ – Batominovski Oct 13 '19 at 14:51
  • $\begingroup$ @Batominovski thanks will fix. $\endgroup$ – Ninad Munshi Oct 13 '19 at 14:53
  • $\begingroup$ I see, ... Wolfram Alpha does not have anything for that integral either. Thanks! $\endgroup$ – Dhvanit Oct 13 '19 at 14:56

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