Solve the boundary value problem y^((4)) + 4y = 0 , with…

Solve the boundary value problem $y^{(4)} + 4y = 0$, with $y(0)=0,\qquad y'(0)=1,\qquad y(x)\ \text{bounded (in fact}\ \to 0\text{) as}\ x\to+\infty$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^4+4=0.$

Factor with the Sophie Germain identity $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$, with $a=r,\ b=1$: $r^4+4=(r^2+2r+2)(r^2-2r+2)=0.$

From $r^2+2r+2=0$: $r=-1\pm i$. From $r^2-2r+2=0$: $r=1\pm i$. Four distinct complex roots.

Step 2: General solution

\[ y = e^{x}\bigl(C_1\cos x+C_2\sin x\bigr) + e^{-x}\bigl(C_3\cos x+C_4\sin x\bigr). \tag{1} \]

Step 3: Apply the boundedness condition at infinity

As $x\to+\infty$, $e^{x}\bigl(C_1\cos x+C_2\sin x\bigr)$ oscillates with unboundedly growing amplitude unless $C_1=C_2=0$; the $e^{-x}$ part always decays to $0$. So boundedness (and the stronger condition $y(\infty)=0$) forces $C_1=C_2=0$, leaving:

\[ y = e^{-x}\bigl(C_3\cos x+C_4\sin x\bigr). \tag{2} \]

Differentiating (2): $y' = e^{-x}\bigl[(C_4-C_3)\cos x-(C_3+C_4)\sin x\bigr].$

Step 4: Apply the conditions at $x=0$

$y(0)=0$: $C_3=0$. $\ y'(0)=1$: $C_4-C_3=1\Rightarrow C_4=1$.

Step 5: Particular solution

The solution is recorded in the result below.

Step 6: Verification

Differentiate successively:

\[ y'=e^{-x}(\cos x-\sin x),\quad y''=-2e^{-x}\cos x,\quad y'''=2e^{-x}(\cos x+\sin x),\quad y^{(4)}=-4e^{-x}\sin x. \] So $y^{(4)}+4y=-4e^{-x}\sin x+4e^{-x}\sin x=0\ \checkmark$. Also $y(0)=0\ \checkmark$, $y'(0)=e^{0}(\cos 0-\sin 0)=1\ \checkmark$, and $\displaystyle\lim_{x\to\infty}e^{-x}\sin x=0\ \checkmark$ since $|e^{-x}\sin x|\le e^{-x}\to0$.