Practice: Second-Order Linear Differential Equations

(a) The characteristic equation. With $y=e^{rx}$ the derivatives are $y^{\prime }=r{e}^{rx}$ and $y^{″}=r^2{e}^{rx}$. Substituting:

The exponential is never zero, so it divides out and

The differential equation has become a quadratic. The exponential is the right trial precisely because differentiating it returns the same function times a constant, so every term acquires the same factor and it cancels. Why constants are required. If $p$ or $q$ depends on $x$, the same substitution gives $(r^2+p(x)r+q(x))e^{rx}=0$, whose bracket still contains $x$. No constant $r$ makes that vanish identically, so there is no characteristic polynomial to solve. For instance $x^2{y}^{″}+x{y}^{\prime }-y=0$ admits no exponential solution of this kind; its solutions are $y=x$ and $y=1/x$. (b) $y^{″}-5y^{\prime }+6y=0$, $y(0)=1$, $y^{\prime }(0)=0$. Characteristic equation $r^2-5r+6=0$, discriminant , so the roots are distinct and real:

confirmed by the factorisation $(r-2)(r-3)$. Hence $y=A{e}^{2x}+B{e}^{3x}$. Differentiate before applying the second condition: $y^{\prime }=2A{e}^{2x}+3B{e}^{3x}$. At $x=0$ every exponential is 1, so the conditions read

Substituting $A=1-B$ into the second: $2-2B+3B=0$, so $B=-2$ and $A=3$:

Verify the conditions. $y(0)=3-2=1$ ✓; $y^{\prime }(0)=6-6=0$ ✓ (numerically $-0.000000$). Verify the equation. $y^{\prime }=6e^{2x}-6e^{3x}$ and $y^{″}=12e^{2x}-18e^{3x}$, so

Each coefficient vanishes separately, as it must, since $e^{2x}$ and $e^{3x}$ are independent. Numerically the residuals are $1.7\times {10}^{-6}$, $4.0\times {10}^{-7}$, $-8.5\times {10}^{-5}$ at $x=0,0.5,1$, which is finite-difference error. Behaviour. Although $y^{\prime }(0)=0$, the solution falls away: $y(1)=-18.00$, $y(2)=-643.06$. The larger root dominates far from the origin, and its coefficient here is negative. (c) $y^{″}+6y^{\prime }+9y=0$, $y(0)=2$, $y^{\prime }(0)=1$. Characteristic equation $r^2+6r+9=0$, discriminant $\mathrm{\Delta }=36-36=0$. The root $r=-3$ repeats, since $r^2+6r+9=(r+3{)}^2$, so the general solution carries the $x$ factor:

Differentiate with the product rule, keeping both terms:

At $x=0$: $y(0)=A=2$, and $y^{\prime }(0)=B-3A=1$, so $B=1+6=7$:

Verify. $y(0)=2$ ✓ and $y^{\prime }(0)=7-6=1$ ✓. Dropping the $-3A$ term from the product rule would give $B=1$, the most common slip here. (d) Why the collapsed family fails. The two terms are not independent, being the same function, so the expression collapses:

Two symbols appear on the page but only one degree of freedom exists. Both terms genuinely solve the equation; the error is in the count of free constants, not in the terms. Why no choice works. From $y=C{e}^{2x}$: $y(0)=C$ and $y^{\prime }=2C{e}^{2x}$, so $y^{\prime }(0)=2C$. Therefore every member of this family satisfies $y^{\prime }(0)=2y(0)$: the slope at the origin is pinned to twice the value there. The conditions demand $y(0)=1$ and $y^{\prime }(0)=0$, which would require $0=2$. The system is unsatisfiable, and no cleverness in choosing $A$ and $B$ can repair it, because the general solution was too small to begin with. The correct solution. The repeated root $r=2$ needs the independent second solution $x{e}^{2x}$, whose ratio to $e^{2x}$ is $x$ rather than a constant. So $y=(A+Bx)e^{2x}$, giving $y(0)=A=1$ and $y^{\prime }(0)=B+2A=0$, hence $B=-2$ and

Why the $x$ factor works. Substituting $y=x{e}^{rx}$ gives $y^{\prime }=(1+rx)e^{rx}$ and $y^{″}=(2r+r^{2}x)e^{rx}$, and with $p=-2r$, $q=r^2$, which hold only for a double root, the substitution yields $2r+r^{2}x-2r-2r^{2}x+r^{2}x=0$ identically. The repair is specific to the repeated case, not a general-purpose trick. (e) Complex roots, and the damping. Characteristic equation $r^2+2r+5=0$, discriminant :

so $\alpha =-1$, $\beta =2$ and $y=e^{-x}(A\cos 2x+B\sin 2x)$. Differentiating the envelope and the bracket together:

At $x=0$, where $\cos 0=1$ and $\sin 0=0$: $y(0)=A=3$, and $y^{\prime }(0)=-A+2B=-1$, so $2B=2$ and $B=1$:

Verify. $-A+2B=-3+2=-1$ ✓; residuals below $6\times {10}^{-6}$ at $x=0,0.5,1$ ✓. Values $y(0)=3$, $y(1)=-0.124764$, $y(2)=-0.367805$, $y(3)=0.129501$, $y(5)=-0.020626$, so the sign alternates while the magnitude shrinks. Why no $i$ survives. Real coefficients force the roots into a conjugate pair, and Euler's identity expands the corresponding exponentials as

Adding the pair and halving gives $e^{\alpha x}\cos \beta x$; subtracting and dividing by $2i$ gives $e^{\alpha x}\sin \beta x$. Both are real and both solve the equation, so real combinations of them exhaust the real solutions. The complex numbers appeared in the working and cancelled from the answer, the same pattern real-coefficient polynomials show generally. The damping. For $y^{″}+c{y}^{\prime }+4y=0$ the discriminant is $c^2-16$: | $c$ | $\mathrm{\Delta }$ | Case | Behaviour |
|---|---|---|---|
| $2$ | | complex | under-damped, oscillating while it decays |
| $4$ | $16-16=0$ | repeated | critically damped, fastest return without overshoot |
| $5$ | | distinct real | over-damped, creeps back slowly | So $c=2$ gives oscillation and $c=4$ returns to rest fastest without overshoot. The critical case is exactly the repeated root of part (d), which a collapsed family would have made unsolvable, and the case door closers and meter needles are tuned to. All three decay, because every root has negative real part; that sign, and nothing else, is the stability question.