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Consider the equation $$\frac{d^2y}{dx^2}+\frac1{x}\left(\frac{dy}{dx}\right)^2-\frac{dy}{dx}=0$$ subject to boundary conditions $y(0)=0\,\, y(1)=1$.

How do we approach this problem. Specifically, the degree of equation being two refuses all attempts to solve it. Is series method the only approach?Are there direct solutions? Does substitution help, or is it similar to some of Clairaut's or Riccat's equation, or the Sturm-Liouville problems? Thanks beforehand.

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    $\begingroup$ Set $y'=z(x)$ and you get a first order ode $\endgroup$ – Yuriy S Jun 23 at 18:17
  • $\begingroup$ @YuriyS but then the resulting equation will be $z'+\frac{z^2}{x}-z=0$, which is not a first order ODE, as the dependency between $z$ and $x$ is not immediate, right? $\endgroup$ – vidyarthi Jun 23 at 18:29
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    $\begingroup$ vidyarthi, why would you think so? we don't need to know the "history" between $z$ and $x$, as long as we can claim that $z=z(x)$ is a function, we can treat it as a first order ODE $\endgroup$ – Yuriy S Jun 23 at 18:30
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Setting $y'=z(x)$ we get:

$$z'+\frac{1}{x} z^2-z=0$$

Let's set $z(x)=\frac{1}{u(x)}$, then the equation becomes:

$$-\frac{u'}{u^2}+\frac{1}{x u^2}-\frac{1}{u}=0$$

Multiplying by $-u^2$:

$$u'+u-\frac{1}{x}=0$$

We have a simple linear ODE.

To apply the boundary conditions we should get back to the original function after solving for $u$.

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  • $\begingroup$ but, again $z$ and $x$ are not related like $z$ and $y$ right?, How could $z$ be a function of $x$? $\endgroup$ – vidyarthi Jun 23 at 18:31
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    $\begingroup$ @vidyarthi, it is. Why wouldn't it be? I suggest you read a calculus I textbook, especially the definitions of a function and a derivative. Main point is: a derivative is also a function. Example: $\sin x$ is a function. We have $(\sin x)'=\cos x$ which is also a function. I don't see a problem here $\endgroup$ – Yuriy S Jun 23 at 18:33
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    $\begingroup$ @vidyarthi, just don't forget, that in this case you first find a function $u(x)$ depending on some constant $C_1$ (it's a first order ODE), then you write $y'(x)=\frac{1}{u(x)}$ and so you know that $y(x)=\int \frac{dx}{u(x)}+C_2$ where $C_1$ and $C_2$ are determined from boundary conditions $\endgroup$ – Yuriy S Jun 23 at 18:37
  • $\begingroup$ so by this method, several such nasty equations can be solved, thank you! never thought that we could solve an equation like $y''+f(x)(y')^n=0$ easily right $\endgroup$ – vidyarthi Jun 23 at 18:39

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