Hello,

While trying to solve the O.D.E. $\displaystyle (D^2+4I)y=\cosh{2x}$

I get a particular solution of $\displaystyle \frac{1}{8} \cosh{2x}$.

But the text book gives $\displaystyle \frac{x}{8} \cosh{2x}$ as a particular solution.

If I plug the particular solution from the text book back into the LHS of the ODE I get $\displaystyle x\cosh{2x}+\frac{1}{2}\sinh{2x}$ which I don't think can be manipulated to equal the RHS.

I can't see what I am missing.

Any help very appreciated.

Scott.