Why is this not a general solution to this nonlinear DE?

a) We can use reduction of order

$$p=y'\tag{1}$$

$$p'=y''\tag{2}$$

The DE becomes

$$p'+p^2=0\tag{3}$$

$$\frac{1}{p^2}p'=-1\tag{4}$$

This last step contains the assumption that ##p^2=y'^2\neq 0##.

$$-\left (\frac{1}{p(x)}-\frac{1}{p(x_i)}\right )=-(x-x_i)\tag{5}$$

$$p(x)=\frac{p(x_i)}{1+p(x_i)(x-x_i)}=y'(x)\tag{6}$$

$$y'(x)=\frac{y'(x_i)}{1+y'(x_i)(x-x_i)}\tag{7}$$

$$=\frac{1}{\frac{1}{y'(x_i)}+x-x_i}\tag{8}$$

$$y(x)=\ln{\left ( \frac{1}{y'(x_i)}+x-x_i \right )} + C\tag{9}$$

b) We can easily verify that ##1## and ##\log{x}## are solutions and that they are linearly independent (their ratio is not constant).

My question is about why ##c_1+c_2\log{x}## is not the general solution.

A few initial observations and mini-questions.

First of all, when we say "general solution" are we implicitly saying "general solution on interval ##[a,b]##"?

That is, can there be a general solution on ##[a_1,b_1]## and different general solution on ##[a_2,b_2]##?

From (9) we can see that there are solutions that are defined for certain intervals with ##x<0##.

Here is one argument I came up with to show that ##c_1+c_2\log{x}## is not a general solution.

Suppose we have two solutions ##y_1## and ##y_2##.

Is ##y_1+y_2## a solution?

$$(y_1''+y_2'')+(y_1'+y_2')^2=(y_1''+y_1'^2)+(y_2''+y_2'^2)+2y_1'y_2'\tag{10}$$

$$=2y_1'y_2'\tag{11}$$

Thus, ##y_1+y_2## is a solution only if ##y_1'y_2'=0##.

This is not always the case. For example, ##y_1(x)=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}## are both solutions but ##y_1'y_2'=\frac{1}{x(1+x)}\neq 0##.

If ##c_1+c_2\ln{x}## were a general solution then we'd have

$$y_1=a_1+a_2\ln{x}\tag{12}$$

$$y_2=b_1+b_2\ln{x}\tag{13}$$

and

$$y_1+y_2=(a_1+b_1)+(a_2+b_2)\ln{x}\tag{14}$$

would be another solution. But our example shows that this isn't true for ##y_1=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}##.

zenterix said:

My question is about why ##c_1+c_2\log{x}## is not the general solution.

Your differential equation is nonlinear. Why should we expect it to be solved by an arbitrary linear combination of two independent solutions? According to your general solution (9) (slightly rewritten), only the specific linear combination ##C_{1}+\log\left(x+C_{2}\right)## solves it.

renormalize said:

Your differential equation is nonlinear. Why should we expect it to be solved by an arbitrary linear combination of two independent solutions? According to your general solution (9) (slightly rewritten), only the specific linear combination ##C_{1}+\log\left(x+C_{2}\right)## solves it.

I don't know what to expect from nonlinear equations at this point. That's why I needed to construct an argument to show this.

renormalize said:

According to your general solution (9) (slightly rewritten), only the specific linear combination C1+log⁡(x+C2) solves it.

Indeed. Just noticed that (9) expresses all solutions as a linear combination of ##1## and ##\log{(x+C_2)}## with coefficients ##C_1## and ##1##, for each given choice of ##C_2##.

Thus, we have linear combinations of different solutions (one pair of solutions for each choice of ##C_2##).

Can you write the solution subject to [itex]y(0) = A[/itex], [itex]y'(0) = B[/itex] as a linear combination [itex]Ay_1+ By_2[/itex] where [itex]y_1[/itex] is the solution subject to [itex]y_1(0) = 1[/itex], [itex]y_1'(0) = 0[/itex] and [itex]y_2[/itex] is the solution subject to [itex]y_2(0) = 0[/itex], [itex]y_2'(0) = 1[/itex]? If not, the general solution cannot be said to be a "linear combination" of solutions.

The solution with ##y_1(0)=1## and ##y_1'(0)=0## is the constant function ##y_1(x)=1##.

The solution with ##y_2(0)=0## and ##y_2'(0)=1## is the function ##y_2(x)=\ln{(1+x)}##.

The way you are posing the question is essentially the same as the original question, which used a slightly different ##y_2(x)=\ln{x}##.

The issue is how to show that one cannot write a solution ##y## as a linear combination of ##y_1## and ##y_2##.

My argument purports to show this with a counterexample to the assertion that one can.

zenterix said:

The solution with ##y_1(0)=1## and ##y_1'(0)=0## is the constant function ##y_1(x)=1##.

The solution with ##y_2(0)=0## and ##y_2'(0)=1## is the function ##y_2(x)=\ln{(1+x)}##.

@zenterix You are indeed correct that the functions ##y_{1}\left(x\right)=1,## ##y_{2}\left(x\right)=\log\left(x+1\right),## ##y\left(x\right)=A+B\log\left(x+1\right)## satisfy all of the criteria of @pasmith in post #4. But ##y\left(x\right)## doesn't solve the original differential equation:$$y''\left(x\right)+\left(y'\left(x\right)\right)^{2}=\frac{B\left(B-1\right)}{\left(x+1\right)^{2}}\neq0$$for an arbitrary choice of ##B##. Thus the general solution cannot be written in the form ##A+B\log\left(x+1\right)##. That's the point of post #4.