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I wanted to find a accuracy solution of the following ODE.

$$y'(x)+y(x)=-y(-x).$$

But when I try to use DSolve as follows

DSolve[y'[x] + y[x] == -y[-x], y[x], x]

I get a warning message:

DSolve::litarg: To avoid possible ambiguity, the arguments of the dependent variable in {y[x]+(y^[Prime])[x]==-y[-x]} should literally match the independent variables.

How can I solve this equation with $y(-x)$?

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    $\begingroup$ As the warning message says, the argument to the dependent variable must be exactly the same as in the last parameter. This is like as if you wrote DSolve[ y'[x]+1==y[z],y[x],x] . May be if you explain what you want to write y[-x] in there instead of y[x] it will help. $\endgroup$
    – Nasser
    Oct 15, 2018 at 2:23
  • $\begingroup$ Is that equation even an ODE? I don't think $y(-x)$ is an allowed term in an "ordinary" differential equation. $\endgroup$ Oct 15, 2018 at 15:26

2 Answers 2

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Put g[x] =f[-x], change the sign x in the equation, then we get a system of two equations that has a solution

 DSolve[{f'[x] + f[x] + g[x] == 0, -g'[x] + g[x] + f[x] == 
   0}, {f, g}, x]

Out[]= {{f -> Function[{x}, (1 - x) C[1] - x C[2]], 
  g -> Function[{x}, x C[1] + (1 + x) C[2]]}}

From the condition g[x] =f[-x]we find C[1] = C[2]

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    $\begingroup$ You could also add f'[0] + 2f[0] == 0 as in @AccidentalFourierTransform's answer. $\endgroup$
    – Carl Woll
    Oct 15, 2018 at 4:33
  • $\begingroup$ Thank you. We can also use the condition g[x]=f[-x] $\endgroup$ Oct 15, 2018 at 4:44
  • $\begingroup$ How did you get this: -g'[x] + g[x] + f[x] == 0 ? $\endgroup$
    – user36313
    Jul 31, 2020 at 7:41
  • $\begingroup$ @user36313 If we change sign x->-x then y'[x]->-y'[-x]=-g'[x]. $\endgroup$ Jul 31, 2020 at 23:12
  • $\begingroup$ Thank you, but I still do not get it. If we change x->-x, then by literal substitution it should be y'[x]->y'[-x]. Why is it not so? $\endgroup$
    – user36313
    Aug 2, 2020 at 6:01
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Take $$ y'(x)+y(x)+y(-x)=0 $$

Write the same equation, with $x\to-x$: $$ y'(-x)+y(x)+y(-x)=0 $$ and subtract the two equations: $$ y'(x)-y'(-x)=0 $$

Integrate this equation: $$ y(x)+y(-x)=2y(0) $$ and plug this back into the initial equation: $$ y'(x)+2y(0)=0 $$ with solution $$ y(x)=y(0)(1-2x) $$

It is easy to check that this solves the initial equation:

y'[x] + y[x] == -y[-x] /. y -> ((-2 y[0] # + y[0]) &)
(* True *)

I hope I didn't mess up the algebra though, so please double check everything.

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