Hello,
I’m having tremendous troubles getting the solution for a differential equation that should be satisfied when the argument of the function is (u-t), it is important that it is at (u-t) since the equation has other functions of t that should affect the solution if say I use another variable. No matter what I do I get the error that (u-t) can’t be used as a variable and I don’t know how to get around this. Any help in this area would be highly appreciated it! (Note: I can find a solution for using t as a variable but when replacing by (u-t) the function found with just t as a variable is no longer a solution)
?A[t_]:=?A[t]=?D^2/(t+?D^2/?A);
?B[t_]:=?B[t]=?D^2/(t+?D^2/?B);
FC[t_]:=FC[t]=-2 H[g] (?B[t]/?D^2-(x (?B[t]-?A[t]))/?D^2)
+((x (x-1))/(2 ?D^2))+2 H[g]^2 ((?B[t]-?A[t])^2/?D^2)+D[H[g],g];
sol=DSolve[FC[t]==0,H[g],g]
Just to follow up, if I change the last like to include only H[t] I get the error:
DSolve::litarg: To avoid possible ambiguity, the arguments of the dependent variable in {((-1+x) x)/(2 \[Sigma]D^2)-2 (1/(t+Power[<<2>>] Power[<<2>>])-(x (-Power[<<2>>] Power[<<2>>]+Power[<<2>>] Power[<<2>>]))/\[Sigma]D^2) H[-t+u]+(2 (-(\[Sigma]D^2/Plus[<<2>>])+\[Sigma]D^2/Plus[<<2>>])^2 H[-t+u]^2)/\[Sigma]D^2-(H^\[Prime])[-t+u]==0} should literally match the independent variables.
I am assuming I can substitute t->u-tt, solve the equation in tt and then reverse the substitution tt->u-t
ga[t_] := sd^2/(t + sd^2/ra);
gb[t_] := sd^2/(t + sd^2/rb);
f = (-2 h[u-t] (gb[t]/sd^2-x (gb[t]-ga[t])/sd^2)+x (x-1)/(2 sd^2)+2 h[u-t]^2 ((gb[t]-ga[t])^2/sd^2)+D[h[u-t],t]==0) /. t->u-tt;
Simplify[(h[tt] /. DSolve[f, h[tt], tt][[1]]) /. tt->u-t]
Please check this carefully to make certain that I have made no mistakes.
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Dear Bill,
Thank you for your answer. Unfortunately it would not work because the other functions, ga[t] and gb[t] have still as a parameter time, so replacing back by (t-u) won’t solve the equation. I ran the code you very kindly provided and called the solution a new function RC, whatever the solution of my original differential equation should make the testRC=zero, and the solution that mathematica gives for the code you provide doesn’t satisfy this 
RC[t_] :=
RC[t] = ( (sd^2 + ra t) (sd^2 +
rb t) (sd^4 + ra rb t (u - C[1]) +
sd^2 (rb (t (1 - x) + x (u - C[1])) +
ra (u + x (t - u) + (-1 + x) C[1])))/(2 (ra -
rb)^2 sd^6 (t - u + C[1])));
testRC = - 2 RC[u - t] (gb[t]/sd^2 - ( x (gb[t] - ga[t]) )/sd^2) + ( (
x (x - 1))/(2 sd^2)) + 2 RC[u - t]^2 ((gb[t] - ga[t])^2/sd^2) +
D[RC[u - t], t];
Simplify[testRC]
Sorry, so I mean that the actual differential equation does have only t for the function ga[t] and gb[t] but is when the input of h[.] is u-t that it should be satisfied.
