Let \(y:\mathbb R\to\mathbb R\) be three times differentiable and satisfy \[ y'''-y=0. \] Suppose \[ \lim_{x\to\infty}y(x)=0. \] Find real numbers \(a,b,c,d\), not all zero, such that \[ ay(0)+by'(0)+cy''(0)=d. \]
Solution
Put \[ \omega\coloneqq\frac{\sqrt3}{2}. \]
1. Every real solution of \(y'''-y=0\) has the form\[ y(x) = A e^x +e^{-x/2}\bigl(B\cos(\omega x)+C\sin(\omega x)\bigr) \]for real constants \(A,B,C\).
The characteristic polynomial is \[ r^3-1 =(r-1)(r^2+r+1), \] whose roots are \[ 1, \qquad -\frac12\pm i\frac{\sqrt3}{2}. \] The standard real solution basis for a constant-coefficient linear ODE therefore gives the displayed form.
2. The hypothesis \(\lim_{x\to\infty}y(x)=0\) forces \(A=0\).
The oscillatory term in step 1 is bounded in absolute value by \[ e^{-x/2}(\abs B+\abs C), \] which tends to zero. If \(A\ne0\), then \(Ae^x\) is unbounded in absolute value as \(x\to\infty\), so the sum cannot tend to zero. Hence \(A=0\).
3. For a decaying solution,\[ y(0)+y'(0)+y''(0)=0. \]
By step 2, \[ y(x) = e^{-x/2}\bigl(B\cos(\omega x)+C\sin(\omega x)\bigr). \] At \(x=0\), \[ y(0)=B, \] and differentiation gives \[ y'(0) =-\frac B2+\frac{\sqrt3}{2}C, \qquad y''(0) =-\frac B2-\frac{\sqrt3}{2}C. \] Adding the three displayed values gives zero.
4. One valid choice is\[ \boxed{a=b=c=1,\qquad d=0}. \]
These four real numbers are not all zero, and step 3 gives \[ ay(0)+by'(0)+cy''(0) =y(0)+y'(0)+y''(0) =0 =d. \]
5. Q.E.D.
Step 4 supplies the required constants.