Problem source

Suppose that
\[{\rm f}''(x)+{\rm f}(-x)=x+3\cos 2x\]
and ${\rm f}(0)=1$, ${\rm f}'(0)=-1$.
If ${\rm g}(x)={\rm f}(x)+{\rm f}(-x)$,  find ${\rm g}(0)$
and show that ${\rm g}'(0)=0$.
Show that
\[{\rm g}''(x)+{\rm g}(x)=6\cos 2x,\]
and hence find ${\rm g}(x)$.
Similarly, if
${\rm h}(x)={\rm f}(x)-{\rm f}(-x)$, find ${\rm h}(x)$
and show that
\[{\rm f}(x)=2\cos x -\cos2x-x.\]

Solution source

\begin{align*}
&& g(0) &= f(0)+f(-0) = 2f(0) = 2 \\
&& g'(x) &= f'(x) - f'(-x) \\
&& g'(0) &= f'(0) - f'(-0) = 0 \\
&& g''(x) &= f''(x) +f''(-x) \\
\Rightarrow && g''(x) + g(x) &= f''(x) +f''(-x) + f(x) + f(-x) \\
&&&= f''(x)+ f(-x)  +f''(-x) + f(x) \\
&&&= x + 3 \cos 2x + (-x + 3 \cos (-2x) ) \\
&&&= 6 \cos 2x \\
\end{align*}

Considering the homogeneous part, we should expected a solution of the form $g(x) = A \sin x + B \cos x$. Seeking an integrating factor of the form $g(x) = C \cos 2x$ we see that $-4C \cos 2x + C \cos 2x = 6 \cos 2x \Rightarrow -3C = 6 \Rightarrow C = -2$. Therefore the general solution is
\begin{align*}
&& g(x) &= A\sin x + B \cos x - 2\cos 2x \\
&& g(0) &= B - 2  = 2\\
&& g'(0) &= A = 0 \\
\Rightarrow && g(x) &= 4\cos x - 2\cos 2x \\
\end{align*}

\begin{align*}
&& h(0) &= f(0) - f(-0) = 0 \\
&& h'(x) &= f'(x) + f'(-x) \\
&& h'(0) &= f'(0) + f'(-0) = -2 \\
&& h''(x) &= f''(x) - f''(-x) \\
\Rightarrow && h''(x) - h(x) &= f''(x) - f''(-x) -( f(x) - f(-x)) \\
&&&= f''(x) +f(-x)- (f''(-x) + f(x))  \\
&&&= x + 3\cos 2x - (-x + 3 \cos(-2x)) \\
&&&= 2x
\end{align*}

Considering the homogeneous part, we should expect a solution of the form $Ae^x + Be^{-x}$. For a specific integral, we can take $-2x$, ie

\begin{align*}
&& h(x) &= Ae^x + Be^{-x} - 2x \\
&& h(0) &= A+B =0 \\
&& h'(0) &= A-B-2 =-2 \\
\Rightarrow && A &=B = 0 \\
\Rightarrow && h(x) &= -2x 
\end{align*}

Therefore

$f(x) = \frac12(f(x) + f(-x)) + \frac12(f(x) -f(-x)) = 2\cos x - \cos 2x -x$