# Problem of the Week #100 - February 24th, 2014

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#### Chris L T521

##### Well-known member
Staff member
We've finally made it to triple digits for the Problem of the Week on MHB!! Thanks again for your support and participation for almost two years now! We wouldn't have made it this far without you! Thanks again to those who participated in last week's POTW! Here's this week's problem (courtesy of MarkFL)!

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Problem: Solve the nonlinear second order ODE $yy^{\prime\prime}=y^3+(y^{\prime})^2$.

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#### Chris L T521

##### Well-known member
Staff member
For this POTW, I'm going to be lenient with grading it; the problem in its own right was challenging. However, there were a couple different approaches and several solutions were in terms of other (correct) functions not found in Mark's suggested solution. After reading through each submission, I've decided that the solutions provided by Ackbach, chisigma, Opalg, Pranav, and Sudharaka are "correct enough" for this problem.

You can find Mark's solution to this problem in all its detail below.

First, we may make the following substitution:

$\displaystyle u=\frac{dy}{dx}\implies \frac{d^2y}{dx^2}=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u\frac{du}{dy}$

and so the given ODE becomes the following system:

$\displaystyle \frac{dy}{dx}=u$

$\displaystyle uy\frac{du}{dy}=y^3+u^2$

The second of these equations may be written as:

$\displaystyle \frac{du}{dy}-\frac{1}{y}u=y^2u^{-1}$ where $u\ne0,\,y\ne0$.

This is a Bernoulli equation, and so mutiplying through by $2u$, we obtain:

$\displaystyle 2u\frac{du}{dy}-\frac{2}{y}u^2=2y^2$

This can also be written as:

$\displaystyle \frac{d}{dy}\left(u^2 \right)-\frac{2}{y}u^2=2y^2$

And this is an equation linear in $u^2$. Thus, we calculate the integrating factor:

$\displaystyle \mu(y)=e^{-2\int\frac{1}{y}\,dy}=y^{-2}$

Hence, the ODE then becomes:

$\displaystyle y^{-2}\frac{d}{dy}\left(u^2 \right)-\frac{2}{y^3}u^2=2$

Recognizing that the left side is the product of a differentiation, we may now write:

$\displaystyle \frac{d}{dy}\left(\frac{u^2}{y^2} \right)=2$

Integrating with respect to $y$, we obtain:

$\displaystyle \int\,d\left(\frac{u^2}{y^2} \right)=2\int\,dy$

$\displaystyle \frac{u^2}{y^2}=2y+c_1$

Thus, we have:

$\displaystyle u=\pm y\sqrt{2y+c_1}$ where $2y+c_1>0$

Now, substituting for $u$ in the first equation of the system we obtained above, we obtain:

$\displaystyle \frac{dy}{dx}=\pm y\sqrt{2y+c_1}$

Separation of variables and integrating gives us:

$\displaystyle \int\frac{1}{y\sqrt{2y+c_1}}\,dy=\pm \int\,dx$

For the integral on the left, it would help for us to develop a formula for:

$\displaystyle \int\frac{1}{w\sqrt{aw+b}}\,dw$

We have three cases to consider:

i) $0<b$

We may rewrite the integrand as follows:

$\displaystyle \frac{1}{w\sqrt{aw+b}}=\frac{1}{w\sqrt{aw+b}}\cdot\frac{\sqrt{aw+b}-\sqrt{b}}{\sqrt{aw+b}-\sqrt{b}}=\frac{1}{\sqrt{b}}\cdot\frac{1}{\sqrt{aw+b}-\sqrt{b}}\cdot\frac{aw-\left(aw+b-\sqrt{b(aw+b)} \right)}{w\sqrt{aw+b}}=$

$\displaystyle \frac{2}{\sqrt{b}}\cdot\frac{\sqrt{w}}{\sqrt{aw+b}-\sqrt{b}}\cdot\frac{\sqrt{w}\dfrac{a}{2\sqrt{aw+b}}-\left(\sqrt{aw+b}-\sqrt{b} \right)\dfrac{1}{2\sqrt{w}}}{w}=$

$\displaystyle \frac{2}{\sqrt{b}}\left(\frac{\sqrt{aw+b}-\sqrt{b}}{\sqrt{w}} \right)^{-1}\frac{d}{dw}\left(\frac{\sqrt{aw+b}-\sqrt{b}}{\sqrt{w}} \right)$

Hence, for this case we have:

$\displaystyle \int\frac{1}{w\sqrt{aw+b}}\,dw=\frac{2}{\sqrt{b}} \ln\left|\frac{\sqrt{aw+b}-\sqrt{b}}{\sqrt{w}} \right|+C$

ii) $b<0$

We may rewrite the integrand as follows:

$\displaystyle \frac{1}{w\sqrt{aw+b}}=a\frac{1}{aw+b-b}\cdot\frac{1}{\sqrt{aw+b}}=\frac{2}{\sqrt{-b}}\cdot\frac{1}{\dfrac{aw+b}{-b}+1}\cdot\frac{1}{\sqrt{-b}}\cdot\frac{a}{2\sqrt{aw+b}}=$

$\displaystyle \frac{d}{dw}\left(\frac{2}{\sqrt{-b}}\tan^{-1}\left(\sqrt{\frac{aw+b}{-b}} \right) \right)$

Hence, for this case we have:

$\displaystyle \int\frac{1}{w\sqrt{aw+b}}\,dw=\frac{2}{\sqrt{-b}}\tan^{-1}\left(\sqrt{\frac{aw+b}{-b}} \right)+C$

iii) $b=0$

$\displaystyle \int\frac{1}{w\sqrt{aw}}\,dw=\frac{1}{\sqrt{a}} \int w^{-\frac{3}{2}}\,dw=\frac{1}{\sqrt{a}}\left(-2w^{-\frac{1}{2}} \right)+C=-\frac{2}{\sqrt{aw}}+C$

And so using these results, we find the solution to the ODE given for the following cases:

i) $0<c_1$

$\displaystyle \frac{2}{\sqrt{c_1}}\ln\left|\frac{\sqrt{2y+c_1}-\sqrt{c_1}}{\sqrt{y}} \right|=\pm x+c_2$

ii) $c_1<0$

$\displaystyle \frac{2}{\sqrt{-c_1}}\tan^{-1}\left(\sqrt{\frac{2y+c_1}{-c_1}} \right)=\pm x+c_2$

iii) $c_1=0$

$\displaystyle \mp\frac{2}{\sqrt{y}}+c_2=\sqrt{2}x\implies y=\frac{4}{\left(x-c_2 \right)^2}$

Note: The function $y\equiv0$ (trivial solution) which we discarded in order to obtain the above solutions also satisfies the given ODE. It is a particular solution not obtainable from the three cases given.

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