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How can we solve $$y'''(t)+a(y''(t))^2+b(y'(t))^3=0$$

Could one make some kind of least common denominator argument to decide possible substitutions? Since the chain rule will come into play, I suppose a substitution both for the variable $t$ and the function $y$ could be useful. Possibly some powers of them?

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    $\begingroup$ You could reduce the order simply by doing $x(t)=y'(t)$. $\endgroup$ – Adrian Keister Jun 25 '18 at 15:04
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    $\begingroup$ Also note that $y(t)=\text{const}$ is a (somewhat trivial) solution. $\endgroup$ – Adrian Keister Jun 25 '18 at 15:05
  • $\begingroup$ This came from the one we solved before, @Dylan I just added a term to make it a bit tougher. $\endgroup$ – mathreadler Jun 25 '18 at 15:14
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As Adrian suggests, let $y'=z$, to get the second order equation $$ z''+a\,(z')^2+b\,z^3=0. $$ Since the independent variable $t$ does not appear explicitly in the equation (I am assuming $a$ and $b$ are constants), we let $$ z'=p,\quad z''=\frac{dp}{dt}=\frac{dp}{dz}\,\frac{dz}{dt}=p\,\frac{dp}{dz}. $$ This gives the first order equation $$ p\,\frac{dp}{dz}+a\,p^2+b\,z^3=0, $$ which written as $$ \frac{dp}{dz}=-a\,p-b\,z^3\,p^{-1} $$ is a Bernoulli equation. To solve it, let $u=p^2$. This gives the linear equation $$ \frac{du}{dz}=-a\,u-b\,z^3. $$ I have not done the calculations, but my impression is that you will not be able to get an explicit solution in terms of elementary functions.

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  • $\begingroup$ I had a feeling it would be something like this, but you were so much faster! +1 $\endgroup$ – mathreadler Jun 25 '18 at 15:15
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    $\begingroup$ Sorry, what just happened...?! So confused... you added new variable and that somehow made it easier? How is $\frac{dp(t)}{dz(t)}=-a\,p(t)-b\,z(t)^3\,p(t)^{-1}$ any better than the original? (I'm not familiar with the solution of Bernoulli's equation.) $\endgroup$ – user541686 Jun 25 '18 at 19:22
  • $\begingroup$ I have edited the answer to explain how to solve the equation. $\endgroup$ – Julián Aguirre Jun 25 '18 at 21:18
  • $\begingroup$ The solution to the last equation is $u=C\exp(-az)+6b/a^4-bz^3/a+3bz^2/a^2-6bz/a^3$ according to MATLAB. This kind of "linear" equations can be solved by combining general solution $u=C\exp(-az)$ of the homogeneous part $\frac{du}{dz}=-a\,u$, and one particular solution of the original equation, a polynomial with similar degree to $-bz^3$. $\endgroup$ – jingyu9575 Jun 26 '18 at 7:10
  • $\begingroup$ @jingyu9575 My comment was about the difficulty of getting an explicit solution the equation $z'=p(z)$. $\endgroup$ – Julián Aguirre Jun 26 '18 at 15:40
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$$y'''(t)+a(y''(t))^2+b(y'(t))^3=0$$ Substitute $z=y'$ $$z''(t)+a(z'(t))^2+bz^3=0$$ Substitute $p=z'$ $$\frac {dp}{dz}p+ap^2+bz^3=0$$ $$\frac 12(p^2)'+ap^2+bz^3=0$$ Finally substitute $w=p^2$ $$\frac 12w'+aw+bz^3=0$$ Bernouilli's equation

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As a more general solution, if you have an equation of the form $$ x''(t) + a(x(t))x'(t)^2+b(x(t)) = 0 $$ then you can make the substitution $f(x) = x'(t)^2$ to arrive at the equation $$ \frac{1}{2}f'(x) + a(x)f(x)+b(x)= 0 $$ Letting $\mu(x) = \exp\left[\int a(x) dx\right]$, we can solve for $f(x)$: $$ f(x) = \mu(x)^{-1}\left(C_1-2\int\mu(x)b(x)dx\right) $$ which can be substituted back for $x(t)$: $$ x' = \mu(x)^{-1/2}\left(C_1-2\int\mu(x)b(x)dx\right)^{1/2} $$ and solved implicitly: $$ C_2 + t - \int \left[ \mu(x)\left(C_1-2\int\mu(x)b(x)dx\right)^{-1} \right]^{1/2} dx = 0 $$

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  • $\begingroup$ but $f'(x) = (x'(t)^2)' =/\text{ chain rule }/= 2x'(t)x''(t) \neq x''(t)$ $\endgroup$ – mathreadler Jun 26 '18 at 4:50
  • $\begingroup$ $df/dx = \frac{df/dt}{dx/dt} = \frac{2x'x''}{x'} = 2x''$. I'm missing a factor of 1/2 there, but otherwise it's correct. I've fixed that error now. See here for substitution: eqworld.ipmnet.ru/en/solutions/ode/ode0344.pdf $\endgroup$ – AlexanderJ93 Jun 26 '18 at 11:11
  • $\begingroup$ I see you switched variables now. Prim notation is confusing when you do that $\endgroup$ – mathreadler Jun 26 '18 at 14:41

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