Worked examples — Case 1 - two distinct real roots

This page is the "no surprises" drill for Case 1. The parent note showed you the machine — Replace, Solve, Superpose. Here we run that machine through every kind of input it can meet, so that when an exam throws you a variation, you have already seen its shape.

Before any algebra, one reminder of the object we are solving:

Here $r_1,r_2$ are the two growth rates; $C_1,C_2$ are free constants set by starting data. Everything below is just this one formula, met in every disguise.

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The scenario matrix

Case 1 always gives two real numbers $r_1,r_2$. What differs is their signs, whether one is zero, whether the problem is dressed as a word problem, and how the constants get fixed. Each row below is a distinct "shape of answer" you must recognise on sight.

Cell	What is special about the roots	Long-term behaviour of $y$	Example
A	Both roots negative ($r_1<0,r_2<0$)	Decays to $0$	Ex 1
B	Both roots positive ($r_1>0,r_2>0$)	Blows up	Ex 2
C	Opposite signs ($r_1>0,r_2<0$)	Generically blows up (one term dominates)	Ex 3
D	One root is zero ($b=0$? no — $c=0$)	Settles to a nonzero constant	Ex 4
E	$b=0$ symmetric roots ($r=\pm k$)	Grows unless the growing part is switched off	Ex 5
F	Initial conditions pin $C_1,C_2$	Depends on data	Ex 6
G	Boundary conditions (values at two $x$'s)	Depends on data	Ex 7
H	Word problem (real-world decay/mixing)	Physical decay	Ex 8
I	Exam twist — leading coefficient $a\ne 1$, needs the quadratic formula	Depends	Ex 9
J	Degenerate boundary — no solution / infinitely many	Watch for failure	Ex 10

The figure below plots the four sign-combinations of roots so you can see what each row does before we compute anything.

Look at the four coloured curves: green (both roots negative) sinks to zero; red (both positive) shoots up; blue (opposite signs) is pulled up by its growing half; yellow (a zero root) flattens onto a constant shelf. Every example below lands on one of these shapes.

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Cell A — both roots negative (pure decay)

1. Replace derivatives by powers of $r$: $r^2+6r+8=0$. Why this step? The ansatz $y=e^{rx}$ makes $y^{\prime \prime }\to r^2{e}^{rx}$, $y^{\prime }\to r{e}^{rx}$; the common factor $e^{rx}\ne 0$ divides out, leaving pure algebra.
2. Check the discriminant: $b^2-4ac=36-32=4>0$. Why this step? Positive discriminant is the entry ticket to Case 1 — two distinct reals.
3. Solve the quadratic: $(r+2)(r+4)=0\Rightarrow r_1=-2,r_2=-4$. Why this step? These are the two growth rates that make the exponential satisfy the ODE.
4. Superpose: $y=C_1{e}^{-2x}+C_2{e}^{-4x}$. Why this step? Superposition plus the need for two free constants in a 2nd-order ODE.

Recall Verify

Take $y=e^{-2x}$: $y^{\prime \prime }+6y^{\prime }+8y=4e^{-2x}-12e^{-2x}+8e^{-2x}=0$ ✔. Both roots negative, so as $x\to \infty$, $y\to 0$ — matches the forecast.

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Cell B — both roots positive (pure growth)

1. Characteristic equation: $r^2-7r+10=0$. Why? Same replacement rule.
2. Discriminant: $49-40=9>0$. Case 1 confirmed. Why? Guarantees two distinct real roots.
3. Factor: $(r-2)(r-5)=0\Rightarrow r_1=2,r_2=5$. Why? Both positive, matching the sum/product prediction.
4. General solution: $y=C_1{e}^{2x}+C_2{e}^{5x}$.

Recall Verify

Sum of roots $=2+5=7=-b/a$ ✔; product $=2\cdot 5=10=c/a$ ✔ (Vieta's checks). Test $y=e^{5x}$: $25e^{5x}-35e^{5x}+10e^{5x}=0$ ✔.

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Cell C — opposite-sign roots (saddle)

1. Characteristic equation: $r^2-r-12=0$.
2. Discriminant: $1+48=49>0$. Why? Case 1.
3. Factor: $(r-4)(r+3)=0\Rightarrow r_1=4,r_2=-3$. Why? One positive ($e^{4x}$ grows), one negative ($e^{-3x}$ decays).
4. General solution: $y=C_1{e}^{4x}+C_2{e}^{-3x}$.

Recall Verify

Product $=4\cdot (-3)=-12=c$ ✔, sum $=4+(-3)=1=-b$ ✔. Test $y=e^{-3x}$: $9e^{-3x}+3e^{-3x}-12e^{-3x}=0$ ✔.

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Cell D — one root is zero (settles to a constant)

1. Characteristic equation: $r^2+3r=0$. Why? Same replacement; $c=0$ leaves no constant term.
2. Factor: $r(r+3)=0\Rightarrow r_1=0,r_2=-3$. Why? These are still two distinct reals, so it is genuinely Case 1 — the zero root is perfectly legal.
3. General solution: $y=C_1{e}^{0\cdot x}+C_2{e}^{-3x}=C_1+C_2{e}^{-3x}$. Why? $e^{0\cdot x}=1$, so the "growth rate zero" solution is the constant function.

Recall Verify

Test $y=1$ (constant): $y^{\prime \prime }+3y^{\prime }=0+0=0$ ✔. As $x\to \infty$, $y\to C_1$, a constant shelf — matches forecast.

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Cell E — symmetric roots $r=\pm k$ ($b=0$)

1. Characteristic equation: $r^2-16=0$.
2. Solve: $r=\pm 4$, i.e. $r_1=4,r_2=-4$. Distinct reals — Case 1. Why? Discriminant $=0-4(1)(-16)=64>0$.
3. General solution: $y=C_1{e}^{4x}+C_2{e}^{-4x}$.

Recall Verify

Test $y=e^{4x}$: $16e^{4x}-16e^{4x}=0$ ✔. Test $y=e^{-4x}$: $16e^{-4x}-16e^{-4x}=0$ ✔.

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Cell F — initial conditions fix the constants

1. Characteristic equation & roots: $r^2-3r+2=(r-1)(r-2)=0\Rightarrow r_1=1,r_2=2$.
2. General solution: $y=C_1{e}^x+C_2{e}^{2x}$.
3. Apply $y(0)=3$: at $x=0$ both exponentials equal $1$, so $C_1+C_2=3$. Why? $e^0=1$ collapses the equation into a plain linear relation.
4. Differentiate: $y^{\prime }=C_1{e}^x+2C_2{e}^{2x}$. Apply $y^{\prime }(0)=4$: $C_1+2C_2=4$. Why? A 2-parameter family needs two conditions to be pinned.
5. Solve the pair: subtract: $(C_1+2C_2)-(C_1+C_2)=4-3\Rightarrow C_2=1$, then $C_1=2$. Why? Two linear equations, two unknowns.
6. Answer: $y=2e^x+e^{2x}$.

Recall Verify

$y(0)=2+1=3$ ✔. $y^{\prime }(0)=2\cdot 1+2\cdot 1=4$ ✔. Plug into ODE: $y^{\prime \prime }-3y^{\prime }+2y=(2e^x+4e^{2x})-3(2e^x+2e^{2x})+2(2e^x+e^{2x})=0$ ✔.

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Cell G — boundary conditions (two different $x$-values)

1. Roots & general solution: $r=\pm 1$, so $y=C_1{e}^x+C_2{e}^{-x}$.
2. Apply $y(0)=1$: $C_1+C_2=1$. Why? Both exponentials are $1$ at $x=0$.
3. Apply $y(\ln 2)=\frac{5}{2}$: $e^{\ln 2}=2$, $e^{-\ln 2}=\frac{1}{2}$, so $2C_1+\frac{1}{2}{C}_2=\frac{5}{2}$. Why? Evaluate the exponentials at $x=\ln 2$ exactly.
4. Solve: from step 2, $C_2=1-C_1$. Substitute: $2C_1+\frac{1}{2}(1-C_1)=\frac{5}{2}\Rightarrow \frac{3}{2}{C}_1=2\Rightarrow C_1=\frac{4}{3}$, then $C_2=-\frac{1}{3}$. Why? Linear elimination again — boundary conditions are still just two linear equations.
5. Answer: $y=\frac{4}{3}{e}^x-\frac{1}{3}{e}^{-x}$.

Recall Verify

$y(0)=\frac{4}{3}-\frac{1}{3}=1$ ✔. $y(\ln 2)=\frac{4}{3}\cdot 2-\frac{1}{3}\cdot \frac{1}{2}=\frac{8}{3}-\frac{1}{6}=\frac{16-1}{6}=\frac{15}{6}=\frac{5}{2}$ ✔.

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Cell H — real-world word problem

1. Characteristic equation: $r^2+5r+4=(r+1)(r+4)=0\Rightarrow r_1=-1,r_2=-4$. Why? Standard replacement.
2. General solution: $y=C_1{e}^{-x}+C_2{e}^{-4x}$.
3. $y(0)=10$: $C_1+C_2=10$. Why? $e^0=1$.
4. $y^{\prime }=-C_1{e}^{-x}-4C_2{e}^{-4x}$, $y^{\prime }(0)=-13$: $-C_1-4C_2=-13\Rightarrow C_1+4C_2=13$. Why? Second condition pins both constants.
5. Solve: subtract step 3: $3C_2=3\Rightarrow C_2=1$, then $C_1=9$.
6. Answer: $y=9e^{-x}+e^{-4x}$. As $x\to \infty$, both terms $\to 0$, so $y\to 0$.

Recall Verify

$y(0)=9+1=10$ ✔. $y^{\prime }(0)=-9-4=-13$ ✔. Units: $y^{\prime }$ shares $y$'s units per second; the rates $-1,-4$ have units $1/\text{s}$ (correct for an exponential of $rx$ with $x$ in seconds). Long-term $y\to 0$ matches physical decay ✔.

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Cell I — exam twist: $a\ne 1$, quadratic formula needed

1. Characteristic equation: $2r^2+5r-3=0$. Why? Keep the coefficient $a=2$; do not divide sloppily.
2. Quadratic formula: $r=\frac{-5\pm \sqrt{25+24}}{4}=\frac{-5\pm 7}{4}$. Why this tool? No obvious integer factoring, so the formula $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ is the general route.
3. Roots: $r_1=\frac{2}{4}=\frac{1}{2},r_2=\frac{-12}{4}=-3$. Why? Discriminant $49>0$, distinct reals — Case 1.
4. General solution: $y=C_1{e}^{x/2}+C_2{e}^{-3x}$.

Recall Verify

Sum $=\frac{1}{2}+(-3)=-\frac{5}{2}=-b/a$ ✔. Product $=\frac{1}{2}\cdot (-3)=-\frac{3}{2}=c/a$ ✔. Test $y=e^{x/2}$: $2\cdot \frac{1}{4}{e}^{x/2}+5\cdot \frac{1}{2}{e}^{x/2}-3e^{x/2}=(\frac{1}{2}+\frac{5}{2}-3)e^{x/2}=0$ ✔.

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Cell J — degenerate boundary problem (a trap)

1. General solution: $y=C_1{e}^x+C_2{e}^{-x}$.
2. Impose ${\lim }_{x\to \infty }y=0$: $e^x\to \infty$, so we must have $C_1=0$. Why? Any nonzero $C_1$ makes $y$ blow up, contradicting the limit.
3. Now $y=C_2{e}^{-x}$. Impose $y(0)=1$: $C_2=1$, so $y=e^{-x}$. A unique solution exists.
4. Second version — impose $y(0)=0$ instead: $C_2=0$, giving $y\equiv 0$. Here the only decaying solution starting at $0$ is the trivial zero function. Why show both? To see that boundary conditions can (a) select a unique nonzero solution, or (b) force the trivial solution — you must check which.

Recall Verify

$y=e^{-x}$: $y^{\prime \prime }-y=e^{-x}-e^{-x}=0$ ✔; $y(0)=1$ ✔; ${\lim }_{x\to \infty }{e}^{-x}=0$ ✔. Version two: $y\equiv 0$ trivially satisfies everything ✔.

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Recall check

Connections

• Case 1 - two distinct real roots — the parent recipe drilled here.
• Characteristic equation — the quadratic behind every example.
• Superposition principle — why every answer is $C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$.
• Wronskian and linear independence — guarantees the two exponentials span all solutions, even when one root is $0$.
• Case 2 repeated real root — the neighbouring case when the discriminant hits $0$.
• Case 3 complex conjugate roots — when the discriminant goes negative.
• Exponential function and its derivative — the reason the whole method works.