Practice Exam 6

Introduction to Differential Equations

Questions10

Time90 min

Points100

RuleVerify every solution

CoversUnit 6

1Classification8 pts

For each, give the order and say whether it is linear:

(a) y″ + 3y′ − 4y = sin t (b) y′ = y² − t (c) ty′ + y = e^t (d) y′y″ = t

Solution

	Order	Linear?	Why
(a)	2	Yes	`y` and derivatives appear to the first power only. The sin `t` is fine — linearity is about `y`, not `t`.
(b)	1	No	`y`² is a nonlinear term.
(c)	1	Yes	A variable coefficient `t` multiplying `y`′ is still linear.
(d)	2	No	A product of two derivatives.

The test: linearity means y and its derivatives appear only to the first power, never multiplied by each other, never inside a function like sin or exp. Coefficients may be any functions of the independent variable. It matters because linear equations have a complete solution theory and nonlinear ones generally do not.

2Separable8 pts

Solve dy/dx = y cos x, y(0) = 2.

Solution

Separate (noting y = 0 is an equilibrium solution, but not ours):

dyy = cos x dx ⇒ ln|y| = sin x + C

y = Ae^{sin x} where A = ±e^C absorbs the constant and the absolute value.

y(0) = 2: since sin 0 = 0, A = 2.

y = 2e^{sin x}

Verify: y′ = 2e^{sin x}cos x = y cos x ✓ and y(0) = 2e⁰ = 2 ✓

Note the solution oscillates between 2e⁻¹ ≈ 0.74 and 2e ≈ 5.44 — bounded and periodic, never reaching zero.

3Separable IVP10 pts

Solve dy/dx = 3x²(1 + y), y(0) = 2.

Solution

dy1 + y = 3x² dx ⇒ ln|1 + y| = x³ + C

1 + y = Ae^x³ ⇒ y = Ae^x³ − 1

y(0) = 2: 2 = A − 1, so A = 3.

y = 3e^x³ − 1

Verify: y′ = 3e^x³(3x²) = 9x²e^x³, and 3x²(1+y) = 3x²(3e^x³) = 9x²e^x³ ✓

Also note: y = −1 is an equilibrium solution (it makes the right side zero). Our solution never reaches it, but a different initial condition would have landed exactly there.

4Integrating factor10 pts

Solve y′ + 2y = e^−x, y(0) = 0.

Solution

Already in standard form, with P = 2:

μ = e^{∫2 dx} = e^2x

e^2xy′ + 2e^2xy = e^2xe^−x = e^x ⇒ (e^2xy)′ = e^x

e^2xy = e^x + C ⇒ y = e^−x + Ce^−2x

y(0) = 0: 0 = 1 + C, so C = −1.

y = e^−x − e^−2x

Recognize the shape: this is exactly the form that came out of the partial-fraction problem in Exam 1, Q10 — a sum of decaying exponentials, starting at zero, rising to a peak, then decaying. It is the step response of a second-order overdamped system, and you have now produced it from both directions.

5Integrating factor · variable coefficient10 pts

Solve xy′ − y = x² for x > 0.

Solution

Standard form first — divide through by x:

y′ − 1xy = x ⇒ P = −1x

μ = e^{∫(−1/x)dx} = e^{−ln x} = 1x

y′x − yx² = 1 ⇒ (yx)′ = 1

yx = x + C ⇒ y = x² + Cx

Verify: y′ = 2x + C, so xy′ − y = 2x² + Cx − x² − Cx = x² ✓

Two places to lose points: not dividing by x first (which gives P = −1 and the wrong μ), and mis-simplifying e^{−ln x} as −x instead of 1/x.

6Second order · complex roots12 pts

Solve y″ − 4y′ + 13y = 0 with y(0) = 1, y′(0) = 2. Is this system stable?

Solution

r² − 4r + 13 = 0 ⇒ r = 4 ± √(16 − 52)2 = 4 ± 6i2 = 2 ± 3i

y = e^2t(C₁cos 3t + C₂sin 3t)

Apply y(0) = 1: C₁ = 1.

Differentiate (product rule — both factors depend on t):

y′ = 2e^2t(C₁cos 3t + C₂sin 3t) + e^2t(−3C₁sin 3t + 3C₂cos 3t)

At t = 0: y′(0) = 2C₁ + 3C₂ = 2. With C₁ = 1, that gives C₂ = 0.

y = e^2tcos 3t

Stability: no. The real part of the roots is +2, so the envelope e^2t grows without bound. The solution oscillates at 3 rad/s with exponentially increasing amplitude — a growing oscillation, which is what an unstable feedback loop does in practice.

The general criterion: a linear system is stable exactly when every characteristic root has a negative real part. Here the y′ coefficient is −4, i.e. negative damping — energy is being pumped in rather than dissipated. In a circuit that requires an active element; passive RLC always has positive R and therefore cannot do this.

7Second order · repeated root10 pts

Solve y″ + 6y′ + 9y = 0, y(0) = 2, y′(0) = 1.

Solution

r² + 6r + 9 = (r+3)² = 0 ⇒ r = −3 (repeated)

y = (C₁ + C₂t)e^−3t The t is mandatory. Writing C₁e^−3t + C₂e^−3t gives only one independent solution, and you cannot satisfy two initial conditions with it.

y(0) = 2 ⇒ C₁ = 2.

y′ = C₂e^−3t − 3(C₁ + C₂t)e^−3t

At t = 0: C₂ − 3(2) = 1, so C₂ = 7.

y = (2 + 7t)e^−3t

Behavior: the linear factor initially wins, so y rises to a maximum before the exponential takes over and drives it to zero. This is the critically damped case — no oscillation, but a single overshoot-free hump. Setting y′ = 0 gives the peak at t = 1/21 s.

8RC discharge · EE12 pts

A 470 µF capacitor charged to 9.0 V discharges through a 22 kΩ resistor. Find (a) the time constant, (b) v(t), (c) the time to fall to 1.0 V, (d) the total energy dissipated in the resistor.

Solution

(a) Time constant:

τ = RC = (22 000)(470×10⁻⁶) = 10.3 s

(b) Natural response — no source, so it decays from the initial value:

v(t) = 9.0 e^−t/10.3 V

(c) Time to 1.0 V:

1.0 = 9.0e^−t/τ ⇒ e^−t/τ = 19 ⇒ t = τ ln 9

t = 10.3(2.197) = 22.7 s

(d) Energy. Everything stored in the capacitor ends up in the resistor:

W = 12CV² = 12(470×10⁻⁶)(9.0)² = 19.0 mJ

Confirm by integrating the power, which is the honest route:

W = ∫₀^∞ v²R dt = 8122 000∫₀^∞ e^−2t/τ dt = 8122 000 · τ2

= 81(10.3)44 000 = 19.0 mJ ✓

The improper integral from Unit 1 and the energy formula from Unit 2 agree. Note the τ/2 — squaring the exponential halves the effective time constant, which is the same reason the energy of a decaying signal settles in half the time the signal does.

9RL circuit · EE10 pts

A 50 mH inductor in series with 100 Ω is connected to 24 V at t = 0. Find (a) τ, (b) the final current, (c) i at t = 1.0 ms, (d) the time to reach 90% of final.

Solution

(a) For an RL circuit the time constant is L/R, not RC:

τ = LR = 0.050100 = 5.0×10⁻⁴ s = 0.50 ms

(b) After a long time the inductor is a short circuit, so the resistor alone sets the current:

I_∞ = 24100 = 0.24 A

i(t) = 0.24(1 − e^{−t/0.5 ms}) A

(c) At t = 1.0 ms, that is exactly 2τ:

i = 0.24(1 − e⁻²) = 0.24(0.8647) = 0.208 A

(d) Time to 90%:

0.90 = 1 − e^−t/τ ⇒ e^−t/τ = 0.10 ⇒ t = τ ln 10 = 1.15 ms

Worth internalizing: t = τ ln 10 ≈ 2.3τ to reach 90% of anything exponential, regardless of the circuit. Likewise ln 2 ≈ 0.69τ for 50%. These two constants save real time.

Contrast with RC: increasing R makes an RC circuit slower but makes an RL circuit faster. The reason is structural: τ = RC versus τ = L/R. Mixing these up is the most common error on RL problems.

10RLC damping · EE10 pts

A series RLC circuit has L = 4.0 mH and C = 25 nF. Find (a) ω₀ and f₀, (b) the resistance for critical damping, (c) classify R = 100 Ω and find the ringing frequency.

Solution

(a) Resonant frequency:

LC = (4.0×10⁻³)(25×10⁻⁹) = 1.0×10⁻¹⁰

ω₀ = 1√(LC) = 11.0×10⁻⁵ = 1.0×10⁵ rad/s

f₀ = ω₀2π = 15.9 kHz

(b) Critical resistance:

R_crit = 2√L/C = 2√4.0×10⁻³ / 25×10⁻⁹ = 2√(1.6×10⁵) = 800 Ω

(c) Classify R = 100 Ω. Well below 800 Ω, so underdamped. Quantitatively:

α = R2L = 1008.0×10⁻³ = 1.25×10⁴ s⁻¹

α = 12 500 < ω₀ = 100 000 ✓ underdamped.

ω_d = √(ω₀² − α²) = √(1.0×10¹⁰ − 1.56×10⁸) = 9.92×10⁴ rad/s

That is 0.8% below ω₀ — the ringing is essentially at the resonant frequency. The damping mostly controls how long it rings, not how fast.

Quality factor summarizes it in one number: Q = ω₀/2α = 100 000/25 000 = 4. Roughly, the oscillation persists for about Q cycles. High Q means a sharp, long-ringing resonance; Q = 0.5 is critical damping.

Scoring

Score	Where you are
90–100	You are ready for a full differential equations course — and for circuits.
75–89	Method selection is fine; the losses are in applying initial conditions. Always verify by substitution.
60–74	Drill the standard-form step and the three characteristic-root cases until they are automatic.
< 60	Re-read the unit, focusing on RC and RL. Those two circuits are 80% of what a first circuits course asks.

Where the points usually go

Not dividing into standard form before computing μ (Q5); omitting the t in the repeated-root solution (Q7); confusing τ = RC with τ = L/R (Q9); and forgetting the product rule when differentiating e^αt(cos + sin) to apply the second initial condition (Q6).

You have finished the course

Worth noticing how much of it converged. Partial fractions from Unit 1 factors the same polynomial that Unit 6 calls the characteristic equation. Euler’s formula from Unit 4 is what turns Unit 6’s complex roots into real oscillations, and what turns Unit 5’s complex numbers into phasors. The improper integral that defines the Laplace transform is Unit 1; the region where it converges is Unit 3.

Every one of those threads ends in the same place: a circuit with a capacitor or an inductor in it. That is not a coincidence in how the course was arranged — it is why this material is required for your degree.