Question About Non-Linear Second Order Differential Equation

[mmmboh]

Hi, I just had my ordinary differential equation final exam today (summer class), the exam was fine. We had a bonus question though, it was a nonlinear second order differential equation, we didn't learn anything about those except for one existence and uniqueness theorem. I am curious if I got it right. The question was, solve:

x²y''-4(y')²=0

Clearly C (constant) is a solution, for the other solution I made an educated guess that y is x², that didn't quite work, but I figured out it was x²/8...so I put
y=x²/8 + C

I know this solution works because I've plugged it in, the thing I am wondering is when I solve it on wolframalpha.com it gives
y[x] = -4 ln(C₁x-4)/(C₁²)-x/C₁+C₂.

I have noticed that these two solutions (mine and wolframs) are linearly independent, but since the equation is nonlinear I don't think combining the two would also be a solution...so I am just wondering if I got the answer right? Also, why doesn't wolfram also give my answer. I mean is it not a completely right answer? I guess the same rules don't apply to nonlinear and linear equations, could there be other answers as well?

Thanks!

 

[Dickfore]

The equation does not contain y explicitly, so you can reduce its order by introducing a new unknown:

$$\begin{array}{l} y' = z \\ y'' = z'$$

The first order equation that you will get is with separable variables and can be integrated.

 

[mmmboh]

Hm yes that worked and I got the answer I got, but what about the answer wolframalpha gives?

 

[mmmboh]

I'm assuming you're talking about the wolfram alpha solution...well, as C₁ approaches zero the equation approaches infinity...I'm not sure..

 

[Dickfore]

If you reduce your equation to the first order equation:

$$x^{2} z' = 4 z^{2}$$
and you integrate this equation by separating variables:
$$\frac{z'}{z^{2}} = \frac{4}{x^{2}}$$

$$\frac{dz}{z^{2}} = \frac{4 \, dx}{x^{2}}$$

$$-\frac{1}{z} = -\frac{4}{x} + C_{1}$$

$$z = \frac{x}{4 - C_{1} \, x}$$

you can see that your solution corresponds to the value $C_{1} = 0$. The reason why this is inconvenient in the general solution is because $C_{1}$ is in the denominator and taking this limit requires that we use the absolute value in the logarithm (remember $\int{dx/x} = \ln|x|$) and adjust the value of $C_{2}$ so that we have an indeterminate form $0/0$ and then take L'Hospital's Rule. The way I am suggesting is more straightforward. So, you get:

$$z = y' = \frac{x}{4}$$

which is precisely what you have found.

 

[mmmboh]

It doesn't seem like that is really the real answer then? You wouldn't know that C1=0. Sorry I am not familiar with non-linear equations, maybe I am applying first order logic that doesn't apply here or something, but does the equation have two linearly independent solutions that you can't combine to make one total solution? If you plug in different values for x in my solutions and in wolframs, you don't get the same answer.

 

[Dickfore]

It doesn't seem like that is really the real answer then? You wouldn't know that C1=0. Sorry I am not familiar with non-linear equations, maybe I am applying first order logic that doesn't apply here or something, but does the equation have two linearly independent solutions that you can't combine to make one total solution? If you plug in different values for x in my solutions and in wolframs, you don't get the same answer.

The red highlighted text addresses two different things: the order of a DE and (non)linearity. Also, the thing you are talking about the two linearly-independent solutions and constructing a general solution by making a linear combination of them is true only for second-order linear DE. As your equation is not linear, this is not the case.

The blue highlighted text also concerns different concepts. When the integrating constants in the general solution attain a particular value, we obtain a particular solution. This is still a function of x. You are not required to take a particular value for the argument of the function x.

As a side note, one particular solution that is "hidden" from the form of the general solution is
$$x = 0$$

To see this, treat y as an argument and x as a dependent variable. The derivatives become:
$$y' = \frac{1}{x'}$$

$$y'' = \frac{dy'}{dy} \, \frac{dy}{dx} = \frac{1}{x'} \, \frac{d}{dy}(\frac{1}{x'}) = -\frac{x''}{(x')^{3}}$$

Substituting these, we get:
$$x^{2} \left(-\frac{x''}{(x')^{3}}\right) - 4 \left(\frac{1}{x'}\right)^{2} = 0$$

$$x^{2} \, x'' + 4 x' = 0$$

Then, $x = 0$ ($x' = x'' = 0$) is an obvious solution.