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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!

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I would try something of the type $f(x)=P(x) \cos(x)$ where $P$ is a fourth degree polynomial... –  N. S. Mar 25 '12 at 21:45
    
That first term is the 4th derivative of y? Would this make things easier for you? If $z=y''+y$, the left side reduces to $z''+z$ –  Mike Mar 25 '12 at 22:09
    
Surely it is the fourth derivative, although perhaps M & W were not consistent in using it. –  GEdgar Mar 25 '12 at 23:34
    
@oenamen: Generally, with a number you are correct that $y^k$ means the $k$th power and $y^{(k)}$ the $k$th derivative; so $y^4$ would mean the $4$th power, and $y^{(4)}$ the fourth derivative. However, when roman numerals are used in the exponent, the meaning, in my experience, is never the power! If you really wanted the fourth power, it would be written $y^4$, not $y^{iv}$. Roman numerals are often used for the derivative, so that the $4$th derivative could be written *either* $y^{(4)}$ or as $y^{iv}$; never as $y^{(iv)}$ or as $y^4$. The fourth power would be $y^4$, not $y^{iv}$ –  Arturo Magidin Mar 25 '12 at 23:42
    
@oenamen: Do you think that is a reasonable, or unreasonable, interpretation? –  anon Mar 26 '12 at 0:38
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2 Answers

up vote 6 down vote accepted

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: $$y_1^{\prime \prime} +y_1 = A\cos x,\qquad y_2^{\prime \prime} +y_2=B\sin x ,\qquad y_3^{\prime \prime} + y_3=\frac{1}{2}x\sin x\; .$$

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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 $$ y=y_h+y_p. $$

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