General solution for $y^{iv}+ 2y''+y=\cos x$

Question

Here is another problem from Mathews and Walker that has given me some trouble.

1-18. Find the general solution of $y^{iv}+ 2y''+y=\cos x$.

Note: Thanks, everyone, for clearing up the interpretation of $y^{iv}$ as the fourth derivative of $y$ and for the clear solutions. I had interpreted $y^{iv}$ as $y^{\sqrt{-1} \ v}$ with $v\in \mathbb{C}$. Of course, this is an awful nonlinear DE!

Answer 1

If you look closely to your ODE's LH side, you discover that: $$y^{(IV)}+2y^{\prime \prime}+y= (y^{\prime \prime} +y)^{\prime \prime} +(y^{\prime \prime}+y)\; .$$ After the substitution $u=y^{\prime \prime} +y$, your ODE rewrites: $$u^{\prime \prime} +u =\cos x\; ,$$ which is a simple second order linear equation and thus can be solved explicitly with ease; in particular, after some computations, you find: $$u(x)=A\ \cos x+ B\ \sin x + \frac{1}{2}\ x\ \sin x\; .$$ Now you can return to the original unknown $y$: the only thing you have to do is to solve the ODE: $$\tag{1} y^{\prime \prime} + y = A\ \cos x+ B\ \sin x + \frac{1}{2}\ x\ \sin x$$ which is again a simple second order linear equation. In order to solve the latter ODE in a clever way, you can observe that $y$ solve (1) iff $y(x)=\bar{y}(x) + y_1(x) + y_2(x) + y_3(x)$ where $\bar{y}$ is the general solution of $\bar{y}^{\prime \prime} + \bar{y} =0$ (which will depend on two arbitrary constants $C,D\in \mathbb{R}$) and $y_1,\ y_2,\ y_3$ are particular solutions of: $$y_1^{\prime \prime} +y_1 = A\cos x,\qquad y_2^{\prime \prime} +y_2=B\sin x ,\qquad y_3^{\prime \prime} + y_3=\frac{1}{2}x\sin x\; .$$

Answer 2

You did not try the easiest way. It is a linear equation with constant coefficients. The characteristic equation is $$ r^4+2\,r^2+1=(r^2+1)^2=0\implies r=\pm i,\text{ roots of multiplicity $2$.} $$ The solution of the homogeneous equation is $$ y_h=C_1\cos x+C_2\sin x+C_3x\cos x+C_4x\sin x. $$ Lookig at the right hans side, we know that there is a particular solution of the complete equation of the form $$ y_p=x^2(A\cos x+B\sin x). $$ $A$ and $B$ are found substituting $y_p$ in the equation. Its general solution is $$ y=y_h+y_p. $$