Worked examples — Case 2 - repeated real root — reduction of order

This page is the drill floor for Case 2 (repeated real root). The parent note proved why the two solutions are $e^{rx}$ and $x{e}^{rx}$. Here we make sure you never meet a version of the problem you haven't already seen solved.

Before anything else, one promise: every symbol you see below was defined in the parent. If you are shaky on where the characteristic equation comes from, glance at Characteristic equation of linear ODEs first.

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The scenario matrix

A "repeated real root" problem is not one problem — it is a family of them. The difficulty hides in the packaging: whether $a=1$ or not, whether $r$ is positive, negative or zero, whether you're asked for a general solution or fed initial/boundary conditions, and whether the whole thing is dressed up as a word problem.

Here is the full grid. Every cell gets at least one fully worked example below.

Cell	What makes it different	The trap it sets	Example
C1	Clean $a=1$, positive root $r>0$	none — warm-up	Ex 1
C2	$a=1$, negative root $r<0$	sign of $r$ in $e^{rx}$	Ex 2
C3	Repeated root $r=0$ (degenerate)	$e^{0x}=1$, so solutions are $1$ and $x$	Ex 3
C4	Leading coefficient $a\ne 1$	must divide/factor carefully for $r=-b/2a$	Ex 4
C5	Initial-value problem (conditions at one point)	applying $y^{\prime }(0)$ correctly	Ex 5
C6	Boundary-value problem (conditions at two points)	solving a $2\times 2$ system	Ex 6
C7	Limiting / coalescing roots viewpoint	seeing $x{e}^{rx}$ emerge as a limit	Ex 7
C8	Word problem (physics: critical damping)	translating words → ODE → root	Ex 8
C9	Exam twist — verify a given function solves it	reverse-engineering $a,b,c$	Ex 9

Figure s01 — read this before the examples. The plot shows the shape of the Case 2 general solution $y=(C_1+C_{2}x)e^{rx}$ for the three sign-classes of the repeated root. The solid black curve is a positive root ($r>0$): it grows without bound. The red curve is a negative root ($r<0$): it rises briefly then decays to zero — this is the critical-damping shape you meet in Ex 5, Ex 6 and Ex 8, and the red arrow points to that monotone decay. The dashed black line is the degenerate root $r=0$: the exponential is $1$, so the solution is just the straight line $1+x$ (Ex 3). One glance tells you which cell of the matrix you are standing in from the sign of $r$ alone.

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Worked Examples

Ex 1 — Cell C1: clean positive root

Forecast: guess the root before reading on — the number $25$ is a perfect square, so what is $r$?

1. Write the characteristic equation. Replace $y^{\prime \prime }\to r^2$, $y^{\prime }\to r$, $y\to 1$: $r^2-10r+25=0.$ Why this step? The trial $y=e^{rx}$ turns derivatives into powers of $r$ (see Characteristic equation of linear ODEs).
2. Check the discriminant. $(-10{)}^2-4(1)(25)=100-100=0$. Why this step? $b^2-4ac=0$ is the signature of a repeated root — it tells us we are in Case 2, not Case 1.
3. Factor as a perfect square. $r^2-10r+25=(r-5{)}^2=0\Rightarrow r=5$ (double). Why this step? Factoring as $(r-r_0{)}^2$ reads off the root with the correct sign, no formula needed.
4. Assemble the two solutions using the Case 2 rule: $e^{5x}$ and $x{e}^{5x}$. Why this step? A 2nd-order ODE needs two independent solutions; the repeated root gives one exponential, and the Case 2 rule supplies the missing partner by tacking on an $x$.
5. General solution: $\boxed{y=(C_1+C_{2}x)e^{5x}}$.

Verify: let $y_2=x{e}^{5x}$. Then $y_2^{\prime }=e^{5x}(1+5x)$, $y_2^{\prime \prime }=e^{5x}(10+25x)$. Substitute: $e^{5x}[(10+25x)-10(1+5x)+25x]=e^{5x}[10+25x-10-50x+25x]=0$. ✓

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Ex 2 — Cell C2: negative repeated root

Forecast: the middle sign is $+$. Will the root be positive or negative?

1. Characteristic equation: $r^2+8r+16=0$. Why this step? Replacing $y^{\prime \prime }\to r^2,y^{\prime }\to r,y\to 1$ under the trial $y=e^{rx}$ converts the ODE into an algebra problem whose roots are the exponents.
2. Discriminant: $8^2-4(16)=64-64=0$ — repeated root confirmed. Why this step? $b^2-4ac=0$ is exactly the condition that puts us in Case 2 (one root, not two).
3. Factor: $(r+4{)}^2=0\Rightarrow r=-4$. Why this step? The $+8r$ produces a root of the opposite sign, $-4$. This is exactly where students misfire — the sign flips.
4. General solution: $\boxed{y=(C_1+C_{2}x)e^{-4x}}$. Why this step? With the repeated root $r=-4$ in hand, the Case 2 pair is $e^{-4x}$ and $x{e}^{-4x}$, combined with two free constants.

Verify: with $y=e^{-4x}$: $y^{\prime \prime }+8y^{\prime }+16y=16e^{-4x}+8(-4)e^{-4x}+16e^{-4x}=(16-32+16)e^{-4x}=0$. ✓

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Ex 3 — Cell C3: the degenerate root $r=0$

Forecast: this looks too simple. What are its two solutions, and does the Case 2 rule still apply?

1. Write it in standard form: $y^{\prime \prime }+0\cdot y^{\prime }+0\cdot y=0$, so $a=1,b=0,c=0$. Why this step? Naming $a,b,c$ explicitly lets us feed them straight into the characteristic equation and discriminant — the machinery expects the equation in this shape.
2. Characteristic equation: $r^2=0\Rightarrow r=0$ (double root). Why this step? $b^2-4ac=0-0=0$, so this is a repeated root — the special value being $r=0$.
3. Apply the Case 2 rule mechanically: solutions are $e^{0x}$ and $x{e}^{0x}$. Why this step? We do NOT skip the rule just because it looks trivial — the pair $\{e^{rx},x{e}^{rx}\}$ holds for every repeated root, including $r=0$.
4. Simplify with $e^{0x}=1$: the solutions are $1$ and $x$. Why this step? $e^{0x}=1$ collapses the exponentials, revealing the two solutions in their plainest form.
5. General solution: $\boxed{y=C_1+C_{2}x}$ — a straight line, exactly what $y^{\prime \prime }=0$ (zero curvature) should give.

Verify: $y=C_1+C_{2}x\Rightarrow y^{\prime \prime }=0$ identically. ✓

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Ex 4 — Cell C4: leading coefficient $a\ne 1$

Forecast: the leading $9$ changes how you extract $r$. Predict $r$ from $r=-b/2a$ before factoring.

1. Characteristic equation: $9r^2-30r+25=0$. Why this step? The same trial $y=e^{rx}$ turns the ODE into a quadratic — but now the leading coefficient is $9$, so we must keep it rather than assume $a=1$.
2. Discriminant: $(-30{)}^2-4(9)(25)=900-900=0$ — repeated root. Why this step? $b^2-4ac=0$ with $a=9$ confirms we are still in Case 2; the non-unit leading coefficient does not change the classification, only the arithmetic.
3. Two safe ways to get $r$:
   • Formula: $r=-\frac{b}{2a}=-\frac{-30}{2\cdot 9}=\frac{30}{18}=\frac{5}{3}$.
   • Perfect square: $9r^2-30r+25=(3r-5{)}^2=0\Rightarrow 3r=5\Rightarrow r=\frac{5}{3}$. Why two ways? They cross-check each other. With $a\ne 1$ the perfect-square factor is $(3r-5)$, not $(r-5)$ — forgetting the $3$ is the classic slip.
4. General solution: $\boxed{y=(C_1+C_{2}x)e^{5x/3}}$. Why this step? With the repeated root $r=\frac{5}{3}$, the Case 2 pair is $e^{5x/3}$ and $x{e}^{5x/3}$; the leading $9$ played no further role once $r$ was found.

Verify: $r=5/3$ must satisfy $9r^2-30r+25=0$: $9\cdot \frac{25}{9}-30\cdot \frac{5}{3}+25=25-50+25=0$. ✓

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Ex 5 — Cell C5: initial-value problem

Forecast: you'll get $y=(C_1+C_{2}x)e^{2x}$. Guess: which condition pins $C_1$, and which pins $C_2$?

1. Characteristic: $r^2-4r+4=(r-2{)}^2=0\Rightarrow r=2$ (double). Why this step? We must solve the general solution first; only then do the conditions have constants to fix.
2. General solution: $y=(C_1+C_{2}x)e^{2x}$. Why this step? The repeated root $r=2$ gives the Case 2 pair; the two free constants are what the two initial conditions will determine.
3. Apply $y(0)=2$. At $x=0$: $e^0=1$ and the $C_{2}x$ term dies. So $y(0)=C_1=2$. Why this step? Evaluating at $x=0$ isolates $C_1$ because $x$ multiplies $C_2$.
4. Differentiate using the product rule: $y^{\prime }=C_2{e}^{2x}+2(C_1+C_{2}x)e^{2x}=e^{2x}[2C_1+C_2+2C_{2}x].$ Why this step? We need $y^{\prime }$ to use the second condition.
5. Apply $y^{\prime }(0)=5$. At $x=0$: $y^{\prime }(0)=2C_1+C_2=5$. With $C_1=2$: $4+C_2=5\Rightarrow C_2=1$. Why this step? The second condition supplies the one remaining equation needed to solve for $C_2$.
6. Answer: $\boxed{y=(2+x)e^{2x}}$.

Verify: $y(0)=2$ ✓. $y^{\prime }(x)=e^{2x}(2\cdot 2+1+2\cdot 1\cdot x)=e^{2x}(5+2x)$, so $y^{\prime }(0)=5$ ✓.

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Ex 6 — Cell C6: boundary-value problem

Forecast: here the two conditions are at different $x$-values. That means a $2\times 2$ system, not a two-step substitution.

1. Characteristic: $r^2+2r+1=(r+1{)}^2=0\Rightarrow r=-1$ (double). Why this step? As always, the general solution must come first before boundary values can pin the constants.
2. General solution: $y=(C_1+C_{2}x)e^{-x}$. Why this step? The repeated root $r=-1$ gives the Case 2 pair $e^{-x},x{e}^{-x}$ with two constants to be fixed by the two boundary conditions.
3. Apply $y(0)=1$: $C_1{e}^0=C_1=1$. Why this step? Same as before — at $x=0$ the $x$-term vanishes, isolating $C_1$.
4. Apply $y(1)=0$: $y(1)=(C_1+C_2)e^{-1}=0$. Since $e^{-1}\ne 0$, we need $C_1+C_2=0\Rightarrow C_2=-C_1=-1$. Why this step? An exponential is never zero, so the bracket must be zero.
5. Answer: $\boxed{y=(1-x)e^{-x}}$.

Verify: $y(0)=1$ ✓. $y(1)=(1-1)e^{-1}=0$ ✓. And $y^{\prime \prime }+2y^{\prime }+y$: with $y=(1-x)e^{-x}$, $y^{\prime }=e^{-x}(-1)-(1-x)e^{-x}=e^{-x}(x-2)$, $y^{\prime \prime }=e^{-x}(1)-(x-2)e^{-x}=e^{-x}(3-x)$; sum $=e^{-x}[(3-x)+2(x-2)+(1-x)]=e^{-x}[3-x+2x-4+1-x]=0$ ✓.

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Ex 7 — Cell C7: watch the roots coalesce

Forecast: the target value is $x{e}^{3x}$ at $x=1$, i.e. $1\cdot e^3=e^3\approx 20.0855$. Watch the difference quotient climb toward it.

1. The setup. Two distinct roots $3$ and $3+\epsilon$ give solutions $e^{3x}$ and $e^{(3+\epsilon )x}$. Their difference quotient is also a solution (linear combination). Why this step? This is Derivation B from the parent — $x{e}^{3x}$ is the derivative of the solution family with respect to the root.
2. Evaluate at $x=1$ for shrinking $\epsilon$: $\frac{e^{3+\epsilon }-e^3}{\epsilon }.$
   • $\epsilon =0.1$: $\frac{e^{3.1}-e^3}{0.1}\approx 21.084$.
   • $\epsilon =0.01$: $\approx 20.186$.
   • $\epsilon =0.001$: $\approx 20.095$. Why this step? Numbers marching toward $e^3\approx 20.0855$ show the limit is $x{e}^{3x}$.
3. Conclusion: the two exponentials fuse into $\{e^{3x},x{e}^{3x}\}$ — the second solution is born from the collision. This is why the factor $x$ is not a lucky guess but a limit.

Figure s02 — what the numbers look like. The red curve is the target limit function $x{e}^{3x}$. The three dashed black curves are the difference quotients $(e^{(3+\epsilon )x}-e^{3x})/\epsilon$ for $\epsilon =0.6,0.3,0.1$. As $\epsilon$ shrinks, each dashed curve hugs the red one more tightly; the red arrow marks where they close in near $x=1$. Watching the dashed curves collapse onto the red limit is the whole point of Derivation B: the second solution $x{e}^{3x}$ literally emerges as two roots slide together.

Verify: exact limit $=\frac{\partial }{\partial r}{e}^r{|}_{r=3}=e^3$, matching the numbers above. ✓

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Ex 8 — Cell C8: critical damping (word problem)

Forecast: with these numbers the discriminant is zero — this is critical damping, the fastest return to rest without oscillating. Guess whether $X(t)$ ever crosses zero.

1. Translate to characteristic equation. With $m=1,c=4,k=4$: $r^2+4r+4=0$. Why this step? $\ddot{X}\to r^2$, $\dot{X}\to r$, $X\to 1$ — the same $e^{rt}$ trial as always, just with time $t$ as the independent variable.
2. Discriminant $=0$: $16-16=0\Rightarrow$ repeated root $(r+2{)}^2=0$, $r=-2$. Why this step? $c^2=4mk$ is precisely the critical damping condition; the maths knows it as a repeated root.
3. General solution: $X(t)=(C_1+C_{2}t)e^{-2t}$. Why this step? The repeated root $r=-2$ gives the Case 2 pair $e^{-2t},t{e}^{-2t}$; the constants will be fixed by the two initial conditions.
4. Apply $X(0)=0.5$: $C_1=0.5$. Why this step? At $t=0$ the $C_{2}t$ term dies and $e^0=1$, isolating $C_1$.
5. Apply $\dot{X}(0)=0$. $\dot{X}=e^{-2t}[C_2-2(C_1+C_{2}t)]=e^{-2t}[C_2-2C_1-2C_{2}t]$. At $t=0$: $C_2-2C_1=0\Rightarrow C_2=2(0.5)=1$. Why this step? "Released from rest" means zero initial velocity, which is exactly the condition $\dot{X}(0)=0$ needed to pin $C_2$.
6. Answer: $\boxed{X(t)=(0.5+t)e^{-2t}\text{m}}$.

Interpretation: since $0.5+t>0$ for all $t\ge 0$ and $e^{-2t}>0$, the mass never crosses zero — it slides monotonically back to equilibrium without overshoot. That's the fingerprint of critical damping (contrast with the oscillating Case 3 complex roots).

Verify: $X(0)=0.5$ ✓. $\dot{X}(t)=e^{-2t}(1-2(0.5+t))=e^{-2t}(-2t)$, so $\dot{X}(0)=0$ ✓. Units: $[t]$ in seconds, $X$ in metres. ✓

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Ex 9 — Cell C9: exam twist (reverse-engineer)

Forecast: the exponent $7x$ screams $r=7$. If it's a repeated root, what must $b$ and $c$ be?

1. Read the root off the exponent. The factor $e^{7x}$ and the presence of an $x$-term $(-2x)$ signal a repeated root $r=7$. Why this step? Only a repeated root produces the $(A+Bx)e^{rx}$ shape — so the packaging of the given function tells us the root and its multiplicity.
2. Repeated root $\Rightarrow$ characteristic is a perfect square: $(r-7{)}^2=r^2-14r+49=0$. Why this step? A double root at $7$ means the quadratic is $(r-7{)}^2$ exactly.
3. Match to $r^2+br+c$: $b=-14$, $c=49$. Why this step? Comparing coefficients of the expanded square with the target polynomial reads off $b$ and $c$ directly.
4. Confirm discriminant $=0$: $b^2-4c=196-196=0$ ✓ — genuinely repeated. Why this step? This closes the loop: the reconstructed coefficients must themselves satisfy the Case 2 condition.
5. Answer: $\boxed{y^{\prime \prime }-14y^{\prime }+49y=0}$, root $r=7$ (double).

Verify: plug $y=(3-2x)e^{7x}$ back. $y^{\prime }=e^{7x}(-2)+7(3-2x)e^{7x}=e^{7x}(19-14x)$. $y^{\prime \prime }=e^{7x}(-14)+7(19-14x)e^{7x}=e^{7x}(119-98x)$. Then $y^{\prime \prime }-14y^{\prime }+49y=e^{7x}[(119-98x)-14(19-14x)+49(3-2x)]=e^{7x}[119-98x-266+196x+147-98x]=e^{7x}[0]=0$ ✓.

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Coverage self-check

Recall Did we hit every cell?

Positive root ::: Ex 1 (C1) Negative root ::: Ex 2 (C2) Degenerate root $r=0$ ::: Ex 3 (C3) Leading coefficient $\ne 1$ ::: Ex 4 (C4) Initial-value problem ::: Ex 5 (C5) Boundary-value problem ::: Ex 6 (C6) Coalescing-roots limit ::: Ex 7 (C7) Physics word problem ::: Ex 8 (C8) Reverse-engineering twist ::: Ex 9 (C9)

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Connections

• Case 2 - repeated real root — reduction of order (index 4.6.12) — the parent theory these examples drill.
• Characteristic equation of linear ODEs — every example starts here.
• Case 1 distinct real roots — contrast when discriminant $>0$.
• Case 3 complex roots — the oscillating cousin of Ex 8 when $c^2<4mk$.
• Reduction of order (general method) — the engine behind the $x{e}^{rx}$ factor.
• Wronskian and linear independence — proves each pair here is genuinely two solutions.
• Higher-order repeated roots — the next tier: multiplicity $m$ gives $x^{m-1}{e}^{rx}$.