# Problem of the Week #272 - Jun 19, 2018

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Find a general solution of the nonlinear differential equation

$$\left(\frac{dy}{dt}\right)^{\!\!2} - y\frac{d^2y}{dt^2} = -1$$
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#### Euge

##### MHB Global Moderator
Staff member
This week’s problem was correctly solved by Opalg . You can read his solution below.

Let $u = \dfrac{dy}{dt}$. Then $\dfrac{d^2y}{dt^2} = \dfrac{du}{dt} = \dfrac{du}{dy}\dfrac{dy}{dt} = u\dfrac{du}{dy}$. Then $\left(\dfrac{dy}{dt}\right)^{\!\!2} - y\dfrac{d^2y}{dt^2} = u^2 - yu\dfrac{du}{dy}$, so the given equation becomes $$yu\frac{du}{dy} = 1 + u^2.$$ Separate the variables and integrate, to get $$\int\frac{u\,du}{1+u^2} = \int\frac{dy}y,$$ $$\frac12\ln(1+u^2) = \ln y + \text{const.}.$$ Exponentiate that, to get $\sqrt{1+u^2} = ay$, for some (nonzero) constant $a$. Then $1+u^2 = a^2y^2$, and $\dfrac{dy}{dt} = u = \sqrt{a^2y^2-1}$. Separate the variables and integrate again: $$\int\frac{dy}{\sqrt{a^2y^2-1}} = \int dt.$$ The left side is a standard integral, equal to $\dfrac1a\ln\left(\sqrt{a^2y^2-1} + ay\right)$, so we get $$\ln\left(\sqrt{a^2y^2-1} + ay\right) = at+b,$$ $$\sqrt{a^2y^2-1} + ay = e^{at+b},$$ $$a^2y^2-1 = \left(e^{at+b} - ay\right)^2,$$ $$-1 = e^{at+b}\left(e^{at+b} - 2ay\right),$$ $$2ay - e^{at+b} = e^{-(at+b)},$$ $$y = \frac1{2a}\left(e^{at+b} + e^{-(at+b)}\right) = \frac1a\cosh(at+b).$$

Since $\sinh^2x - \cosh^2x = -1$, it is almost immediate to check that this satisfies the given differential equation.

There are places in the solution where it would be equally valid to take the negative value of a square root. But that would lead to the same family of solutions $y = \frac1a\cosh(at+b).$

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