DSolve not returning "trivial" solutions

Question

When I enter this

DSolve[y'[x]^2 + y[x]^2 == 1, y[x], x]

the answer I get is

{{y[x] -> -Sin[x - C[1]]}, {y[x] -> Sin[x + C[1]]}}

but the two functions $y(x)=1$ and $y(x)=-1$ which are also solutions to this differential equation aren't mentioned. Is this an oversight or am I missing something?

This question is answered here DSolve not finding solution I expected. The only difference is that there we had z[x]^2 == z'[x]. So it is just a duplicate. ` — Artes, Sep 06 '14 at 16:41
@Artes Can you explicitly show how to get the non-general solutions in this case? — Teake Nutma, Sep 06 '14 at 18:00
@TeakeNutma The question is "Is this an oversight or am I missing something?" and the answer there is in the linked post. This is not "How do I get a special solution?" so I'm not going to play with it. Right? — Artes, Sep 06 '14 at 18:26

score 1 · Answer 1 · answered Sep 06 '14 at 16:52

If you want to include the case where the derivative is zero everywhere (which is not a differential equation), you can use

Join[
 DSolve[y'[x]^2 + y[x]^2 == 1, y[x], x],
 Solve[y[x]^2 == 1, y[x]]]

{{y[x] -> -Sin[x - C[1]]}, {y[x] -> Sin[x + C[1]]}, {y[x] -> -1}, {y[x] -> 1}}

Although, DSolve does not complain about the lack of a derivative in the equation

Join[
 DSolve[y'[x]^2 + y[x]^2 == 1, y[x], x],
 DSolve[y[x]^2 == 1, y[x], x]]

{{y[x] -> -Sin[x - C[1]]}, {y[x] -> Sin[x + C[1]]}, {y[x] -> -1}, {y[x] -> 1}}

% == %%

True

DSolve not returning "trivial" solutions

1 Answers1