Introduction to Differential Equations

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.

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

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

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

Note the solution oscillates between \(2e^{-1} \approx 0.74\) and \(2e \approx 5.44\) — bounded and periodic, never reaching zero.

\(y(0) = 2\): \(2 = A - 1\), so \(A = 3\).

Verify: \(y' = 3e^{x^3}(3x^2) = 9x^2e^{x^3}\), and \(3x^2(1+y) = 3x^2(3e^{x^3}) = 9x^2e^{x^3}\) ✓

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.

Already in standard form, with \(P = 2\):

\(y(0) = 0\): \(0 = 1 + C\), so \(C = -1\).

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.

Standard form first — divide through by \(x\):

Verify: \(y' = 2x + C\), so \(xy' - y = 2x^2 + Cx - x^2 - Cx = x^2\) ✓

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

Apply \(y(0) = 1\): \(C_1 = 1\).

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

At \(t = 0\): \(y'(0) = 2C_1 + 3C_2 = 2\). With \(C_1 = 1\), that gives \(C_2 = 0\).

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.

\(y(0) = 2 \implies C_1 = 2\).

At \(t = 0\): \(C_2 - 3(2) = 1\), so \(C_2 = 7\).

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.

(a) Time constant:

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

(c) Time to 1.0 V:

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

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

The improper integral from Unit 1 and the energy formula from Unit 2 agree. Note the \(\tau/2\) — 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.

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

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

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

(d) Time to 90%:

Worth internalizing: \(t = \tau\ln 10 \approx 2.3\tau\) to reach 90% of anything exponential, regardless of the circuit. Likewise \(\ln 2 \approx 0.69\tau\) 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: \(\tau = RC\) versus \(\tau = L/R\). Mixing these up is the most common error on RL problems.

(a) Resonant frequency:

(b) Critical resistance:

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

\(\alpha = 12\,500 < \omega_0 = 100\,000\) ✓ underdamped.

That is 0.8% below \(\omega_0\) — 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: \(Q = \omega_0/2\alpha = 100\,000/25\,000 = \mathbf{4}\). Roughly, the oscillation persists for about \(Q\) cycles. High \(Q\) means a sharp, long-ringing resonance; \(Q = 0.5\) is critical damping.