# Question:Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument

## Question:Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument

Maple 2017

hello

How I can remove these errors?

Thanks

 >
 (1.1)
 >

 > eq33a:=subs(t=tau/omega,value(subs(u(t)=u(omega*t),eq31g)));
 (2.1)
 (2.2)
 (2.3)
 > temp:=subs(uExpRule,omgRule,convert(eq33a,diff));
 > eq33b:=convert(series(lhs(temp),epsilon,4),polynom)=0;
 > eqEps:=seq(coeff(lhs(eq33b),epsilon,i)=0,i=1..3);

The general solution of the first-order equation, eqEps[1], can be expressed as

 > sol1:=dsolve({eqEps[1],u[1](0)=a,D(u[1])(0)=0},u[1](tau));
 > eq33c:=subs(u[1]=u[2],lhs(eqEps[1])=lhs(eqEps[1]))-subs(sol1,0=lhs(eqEps[2]));

Expanding the right-hand side of eq33c in a Fourier series using trigonometric identities yields

 > eq33c_RHS:=combine(rhs(eq33c));

Eliminating the terms,  and , demands that . Then, the particular solution of eq33c can be expressed as

 > sol2:=combine(subs(_C1=0,_C2=0,dsolve(subs(omega[1]=0,eq33c),u[2](tau))));

Substituting sol1 and sol2 into the third-order equation, eqEps[3], and using the fact that , we obtain

 > eq33d:=subs(u[1]=u[3],lhs(eqEps[1])=lhs(eqEps[1]))-subs(sol1,sol2,omega[1]=0,0=lhs(eqEps[3]));

Expanding the right-hand side of eq33d in a Fourier series using trigonometric identities, we have

 > eq33d_RHS:=combine(rhs(eq33d));

Eliminating the terms that lead to secular terms from eq33d_RHS demands that

 > omg2Rule:=omega[2]=solve(coeff(eq33d_RHS,cos(beta+tau)),omega[2]);
 (2.4)

As discussed above, for a second-order uniform expansion, we do not need to solve for . Combining the first- and second-order solutions, we obtain, to the second approximation, that

 > combine(subs(sol1,sol2,uExpRule));

where

 > tau=subs(omega[1]=0,omg2Rule,subs(omgRule,omega*t));
 (2.5)