Linear Homogeneous Differential Equations — Question 6

Question 6 – Solution

Strategy. Track the quadratic roots and separately check when either merges with the fixed root.

Step 1: Locate the collisions. Besides $r=-1$, the roots are $r_\pm=-\alpha\pm\sqrt{\alpha^2-1}$. They coincide when $\alpha=\pm 1$. Substituting $r=-1$ in the quadratic gives $2-2\alpha$, so coincidence with the fixed root occurs only at $\alpha=1$, producing a triple root $-1$.

Step 2: State a complete basis in each case. For $|\alpha|>1$, all three roots are real and distinct, and $$y=C e^{-x}+A e^{r_+x}+B e^{r_-x}.$$ For $|\alpha|<1$, let $\beta=\sqrt{1-\alpha^2}>0$. Then $$y=C e^{-x}+e^{-\alpha x}(A\cos\beta x+B\sin\beta x).$$ The exceptional cases are $$\begin{array}{ll} \alpha=1:& y=e^{-x}(C+Ax+Bx^2),\\[2pt] \alpha=-1:& y=C e^{-x}+(A+Bx)e^x. \end{array}$$ Each family has three free constants.

Step 3: Classify all forward behavior. If $\alpha>1$, both quadratic roots are negative since $\sqrt{\alpha^2-1}<\alpha$. If $0<\alpha<1$, their real part is $-\alpha<0$. At $\alpha=1$, even the quadratic polynomial factor is dominated by $e^{-x}$. Thus every solution decays for every $\alpha>0$.

At $\alpha=0$, the modes are $e^{-x},\cos x,\sin x$: all are bounded, but not all decay. If $-1<\alpha<0$, the oscillatory amplitude grows exponentially; choosing a cosine and its phase maxima proves unboundedness. At $\alpha=-1$ there are positive-root modes, and for $\alpha<-1$ both quadratic roots are positive. Consequently, $$\boxed{\text{forward bounded}\iff\alpha\ge 0,\qquad \text{forward decaying}\iff\alpha>0.}$$

Step 4: Test the special parameters concretely. The requested examples are $\cos x$ at $\alpha=0$, $x^2e^{-x}$ at $\alpha=1$, and $xe^x$ at $\alpha=-1$. A finite polynomial factor does not defeat exponential decay at a negative root. A positive root already permits growth, and repetition adds further polynomial amplification.