# Differential Equation with y(-x)

Please how can I solve $$y''(x)+y'(-x)=e^x$$

I tried everything I could I can't even find the complementary solution Any help would be gladly appreciated

Put $u(x) = \frac{y(x) + y(-x)}{2}$ and $v(x) = \frac{y(x) - y(-x)}{2}$. So $y(x) = u(x) + v(x)$ and $y(-x) = u(x) - v(x)$. Then your equation is $$\tag{1}u''(x) + v''(x) + v'(x) - u'(x) = e^x.$$ Now you've eliminated the $-x$ from the equation at the cost of adding an extra dependent variable, however you know something about the parity of $u$ and $v$: $u$ is even and $v$ is odd so one possible way to solve is to split the equation into even and odd parts. Since $u$ is even, $u'$ is odd and $u''$ is even. Likewise since $v$ is odd, $v'$ is even and $v''$ is odd. Thus if we can solve the coupled system $$\tag{2} u''(x) + v'(x) = \cosh(x)$$ and $$\tag{3}v''(x) -u'(x) = \sinh(x)$$ then we'll have solution to $(1)$. Differentiating $(3)$, gives $$u''(x)=v'''(x) - \cosh(x)$$ whence, inserting this into $(2)$ gives $$v'''(x) + v'(x) = 2\cosh(x).$$ This has solution $$v(x) = \sinh(x) + A + B\sin(x) + C\cos(x)$$ and since we know $v$ is odd, we must have $A,C = 0$ so $$\boxed{v(x) = \sinh(x) + B\sin(x),}$$ for some constant $B$. Similarly, differentiating $(2)$ gives $$v''(x) = -u'''(x) + \sinh(x)$$ and inserting this into $(3)$ yields $$u'''(x) + u'(x) = 0$$ which (once we've accounted for the fact that $u$ is even) has solution $$\boxed{u(x) = A + C\cos(x)}$$ for some new constants $A,C$. Then $y(x) = u(x) + v(x)$ yields the solution $$y(x) = \sinh(x) + A + B\sin(x) + C\cos(x).$$ At first this looks like too many linearly independent homogeneous solutions, but if we plug our solution for $u,v$ into $(2)$, we find $$-C\cos(x) + B\cos(x) = 0$$ so $B= C$. Thus our final solution is $$\boxed{y(x) = \sinh(x) + A + B(\sin(x) + \cos(x)),}$$ for arbitrary constants $A,B$.