In the equation y'' + 2y' + (λ^2+1)y = 0 , determine λ so…

In the equation $y'' + 2y' + (\lambda^2+1)y = 0$, determine $\lambda$ so that some (nontrivial) solution vanishes at $x=0$ and at $x=2\pi$, with exactly two further zeros strictly between $0$ and $2\pi$.

Step 1: Characteristic roots and the general solution

The characteristic equation is $r^2+2r+(\lambda^2+1)=0$, so

\[ r = -1 \pm \sqrt{1-(\lambda^2+1)} = -1 \pm \sqrt{-\lambda^2} = -1 \pm i\lambda \qquad (\lambda\neq0). \]

These are complex conjugate roots, giving

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

Since $e^{-x}>0$ for every $x$, the zeros of $y$ coincide exactly with the zeros of $f(x)=C_1\cos\lambda x+C_2\sin\lambda x$; the exponential factor only damps the amplitude, it never itself vanishes or shifts a zero.

Step 2: Imposing $y(0)=0$

At $x=0$: $y(0)=e^0(C_1\cdot1+C_2\cdot0)=C_1=0$. So the solution reduces to

\[ y = C_2\,e^{-x}\sin(\lambda x), \]

and for a nontrivial solution we need $C_2\neq0$.

Step 3: Imposing $y(2\pi)=0$

At $x=2\pi$: $y(2\pi)=C_2e^{-2\pi}\sin(2\pi\lambda)=0$. Since $C_2\neq0$ and $e^{-2\pi}\neq0$, this forces

\[ \sin(2\pi\lambda) = 0 \;\Longrightarrow\; 2\pi\lambda = n\pi \;\Longrightarrow\; \lambda = \frac{n}{2}, \qquad n\in\mathbb{Z},\ n\neq0. \]

Since replacing $\lambda$ by $-\lambda$ only flips the sign of $\sin\lambda x$ (equivalently, replaces $C_2$ by $-C_2$) and gives the same family of solution curves, it suffices to take $\lambda>0$, i.e. $\lambda=n/2$ for $n=1,2,3,\dots$.

Step 4: Counting the interior zeros

With $\lambda=n/2$, the zeros of $y=C_2e^{-x}\sin\!\left(\dfrac{n}{2}x\right)$ occur where

\[ \frac{n}{2}x = k\pi \;\Longrightarrow\; x = \frac{2k\pi}{n}, \qquad k=0,1,2,\dots,n. \]

The endpoints $x=0$ ($k=0$) and $x=2\pi$ ($k=n$) are the prescribed boundary zeros; the values $k=1,\dots,n-1$ give exactly

\[ n-1 \]

zeros strictly inside $(0,2\pi)$.

Step 5: Selecting $\lambda$ for exactly two interior zeros

We require $n-1=2$, so $n=3$, and hence

\[ \lambda = \frac{n}{2} = \frac32. \]

The corresponding solution is $y=C_2e^{-x}\sin\!\left(\dfrac{3x}{2}\right)$, with its two interior zeros at $x=\dfrac{2\pi}{3}$ and $x=\dfrac{4\pi}{3}$ (from $x=2k\pi/3$, $k=1,2$).

Step 6: Verification

With $\lambda=3/2$, the equation becomes $y''+2y'+\dfrac{13}{4}y=0$, and $y=C_2e^{-x}\sin(3x/2)$ is its solution vanishing at $x=0$. Checking the four zero locations directly:

\[ \sin\!\left(\frac32\cdot0\right)=0,\quad \sin\!\left(\frac32\cdot\frac{2\pi}{3}\right)=\sin\pi=0,\quad \sin\!\left(\frac32\cdot\frac{4\pi}{3}\right)=\sin2\pi=0,\quad \sin\!\left(\frac32\cdot2\pi\right)=\sin3\pi=0. \]

All four values $x=0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ 2\pi$ are confirmed zeros, and the middle two are strictly inside $(0,2\pi)$ as required, with no additional zeros in between since $\sin(3x/2)$ completes exactly $3/2$ periods, hence exactly $3$ half-period zero crossings beyond $x=0$, over $x\in[0,2\pi]$.\ \checkmark\)