Question:
$$y''+(y')^2=0$$
Verify that $y_1=1$ and $y_2=\ln(x)$ are linearly independent solutions of the equation on any interval to the right of the origin. Is $c_1+c_2\ln(x)$ the general solution. If not, why not?
My attempt:
Direct substitution into the DE shows both $y_1$ and $y_2$ are solutions.
To show they are linearly independent either by ratio or by the Wronskian:
$$W(1,\ln(x)) = \frac{1}{x} \neq 0 \space\space\space\space\space \forall x>0$$
So since both solutions are linearly independent then the general solution should be: $$y=c_1+c_2\ln(x)$$
However, substituting this into the original differential equation, I get:
$$-\frac{c_2}{x^2}+\frac{c_2^2}{x^2} = \frac{c_2^2-c_2}{x^2} \neq 0$$
So why does not it satisfy the original equation?
Moreover, solving the DE using reduction of order, I get: $$y=\ln(x-c_1)+c_2$$ as the general solution (which works). So how to answer the "why not question"?
Best Answer
The equation is not linear, so the superposition principle does not apply. Linear independence is thus the wrong property. One could explore for functional independence, but there the search space of multivariate functions is much larger than the space of coefficient tuples.
Multiply with $e^y$ to get $$(e^y)''=0,$$ which has the solutions $$y(x)=\ln(c_1x+c_0).$$ You can find the claimed solutions among them.