Differential Equations -- Diagnostic Tests

This topic explores fundamental concepts that shape our understanding of the world.

Question: Solve $\frac{dy}{dx} = \frac{x}{y}$ given that $y = 2$ when $x = 1$. (b) Solve $\frac{dy}{dx} = xy$ given that $y = 1$ when $x = 0$. (c) Solve $\frac{dy}{dx} = \frac{y^2 + 1}{y}$ given that $y = 0$ when $x = 0$. (d) Explain what is meant by a “particular solution” versus a “general solution”.

Solution:

(a) $y\,dy = x\,dx$. $\frac{y^2}{2} = \frac{x^2}{2} + C$. $y^2 = x^2 + C$. At $(1,2)$: 4 = 1 + C$$C = 3. y^2 = x^2 + 3$$y = \sqrt{x^2 + 3} (taking positive root).

(b) $\frac{dy}{y} = x\,dx$. $\ln|y| = \frac{x^2}{2} + C$. $y = Ae^{x^2/2}$. At $(0,1)$: $1 = A$So $y = e^{x^2/2}$.

(c) $\frac{y}{y^2 + 1}\,dy = dx$. Integrating: $\frac{1}{2}\ln(y^2 + 1) = x + C$. $\ln(y^2 + 1) = 2x + C$. At $(0,0)$: $\ln 1 = C = 0$. y^2 + 1 = e^{2x}$$y = \sqrt{e^{2x} - 1}.

(d) The general solution contains an arbitrary constant $C$ and represents the family of all possible solutions. The particular solution is obtained by using an initial/boundary condition to find the value of $C$Giving a single specific solution.

Question: (a) Solve $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$. (b) Solve $\frac{d^2y}{dx^2} + 4y = 0$ given that $y = 1$ and $\frac{dy}{dx} = 0$ when $x = 0$. (c) Solve $\frac{d^2y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$. (d) Explain the significance of the discriminant of the auxiliary equation.

Solution:

(a) Auxiliary equation: $m^2 - 5m + 6 = 0$. $(m-2)(m-3) = 0$. $m = 2, 3$. General solution: $y = Ae^{2x} + Be^{3x}$.

(b) $m^2 + 4 = 0$. $m = \pm 2i$. General solution: $y = A\cos 2x + B\sin 2x$. $y(0) = 1$: $A = 1$. $y"(0) = 0$: 2B = 0$$B = 0. Particular solution: $y = \cos 2x$.

(c) $m^2 + 6m + 9 = 0$. $(m+3)^2 = 0$. $m = -3$ (repeated root). General solution: $y = (A + Bx)e^{-3x}$.

(d) The discriminant $\Delta = b^2 - 4ac$ of the auxiliary equation $am^2 + bm + c = 0$ determines the nature of solutions:

• $\Delta \gt 0$: Two distinct real roots, general solution $y = Ae^{m_1x} + Be^{m_2x}$.
• $\Delta = 0$: One repeated real root, general solution $y = (A + Bx)e^{mx}$.
• $\Delta \lt 0$: Complex conjugate roots $m = \alpha \pm \beta i$General solution $y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$.

Question: (a) Find the particular integral and general solution of $\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = 4$. (b) Solve $\frac{d^2y}{dx^2} + y = 10\cos x$. (c) Solve $\frac{d^2y}{dx^2} - 4y = 2e^{3x}$.

Solution:

(a) CFE: m^2 - 3m + 2 = 0$$m = 1, 2. CF: $y_c = Ae^x + Be^{2x}$. PI: try $y_p = k$. 0 - 0 + 2k = 4$$k = 2. GS: $y = Ae^x + Be^{2x} + 2$.

(b) CFE: m^2 + 1 = 0$$m = \pm i. CF: $y_c = A\cos x + B\sin x$. PI: since $\cos x$ is in the CF, try $y_p = x(C\cos x + D\sin x)$. $y_p' = C\cos x + D\sin x + x(-C\sin x + D\cos x)$. $y_p'' = -C\sin x + D\cos x - C\sin x + D\cos x + x(-C\cos x - D\sin x) = -2C\sin x + 2D\cos x - x(C\cos x + D\sin x)$. $y_p'' + y_p = -2C\sin x + 2D\cos x = 10\cos x$. 2D = 10 \Rightarrow D = 5$$-2C = 0 \Rightarrow C = 0. PI: $y_p = 5x\sin x$. GS: $y = A\cos x + B\sin x + 5x\sin x$.

(c) CFE: m^2 - 4 = 0$$m = \pm 2. CF: $y_c = Ae^{2x} + Be^{-2x}$. PI: try $y_p = ce^{3x}$. $9ce^{3x} - 4ce^{3x} = 2e^{3x}$. 5c = 2$$c = 2/5. GS: $y = Ae^{2x} + Be^{-2x} + \frac{2}{5}e^{3x}$.

Question: A particle of mass $m$ falls under gravity with air resistance proportional to velocity: $m\frac{dv}{dt} = mg - kv$. (a) Solve the DE given $v = 0$ when $t = 0$. (b) Find the terminal velocity. (c) Calculate $v$ when $t = m/k$ as a fraction of terminal velocity. (d) Sketch the velocity-time graph.

Solution:

(a) $\frac{dv}{dt} = g - \frac{k}{m}v$. Let $\alpha = k/m$: $\frac{dv}{dt} + \alpha v = g$. Integrating factor: $e^{\alpha t}$. $\frac{d}{dt}(ve^{\alpha t}) = ge^{\alpha t}$. $ve^{\alpha t} = \frac{g}{\alpha}e^{\alpha t} + C$. $v = \frac{g}{\alpha} + Ce^{-\alpha t} = \frac{mg}{k} + Ce^{-kt/m}$. At $t = 0$: 0 = mg/k + C$$C = -mg/k. $v = \frac{mg}{k}(1 - e^{-kt/m})$.

(b) Terminal velocity: as t \to \infty$$e^{-kt/m} \to 0So $v_{\text{term}} = \frac{mg}{k}$.

(c) At $t = m/k$: $v = \frac{mg}{k}(1 - e^{-1}) = \frac{mg}{k}(1 - 0.368) = 0.632 v_{\text{term}}$. The particle reaches approximately 63.2% of terminal velocity.

(d) The graph starts at $v = 0$ with steep gradient $g$Curves concavely, and asymptotically approaches $v = mg/k$. It is a typical exponential approach curve.

Question: Solve the system: \frac{dx}{dt} = 2x + y$$\frac{dy}{dt} = x + 2yWith x(0) = 1$$y(0) = 0. (a) Write in matrix form. (b) Find eigenvalues and eigenvectors of the coefficient matrix. (c) Solve the system. (d) Calculate $x(1)$ and $y(1)$.

Solution:

(a) $\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$.

(b) From the matrix diagnostics: eigenvalues $\lambda_1 = 3$ (eigenvector $\begin{pmatrix}1\\1\end{pmatrix}$) and $\lambda_2 = 1$ (eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$).

(c) $\begin{pmatrix}x\\y\end{pmatrix} = c_1 e^{3t}\begin{pmatrix}1\\1\end{pmatrix} + c_2 e^{t}\begin{pmatrix}1\\-1\end{pmatrix}$.

$\begin{pmatrix}1\\0\end{pmatrix} = c_1\begin{pmatrix}1\\1\end{pmatrix} + c_2\begin{pmatrix}1\\-1\end{pmatrix}$. $c_1 + c_2 = 1$ and $c_1 - c_2 = 0$So $c_1 = c_2 = 1/2$.

x = \frac{e^{3t} + e^t}{2}$$y = \frac{e^{3t} - e^t}{2}.

(d) $x(1) = \frac{e^3 + e}{2} = \frac{20.086 + 2.718}{2} = 11.40$. $y(1) = \frac{e^3 - e}{2} = \frac{20.086 - 2.718}{2} = 8.68$.

Question: The current $I(t)$ in an RL circuit with inductance $L$ and resistance $R$ satisfies $L\frac{dI}{dt} + RI = V_0\cos(\omega t)$. (a) Solve this DE given $I(0) = 0$. (b) Identify the transient and steady-state components. (c) Find the amplitude of the steady-state current. (d) Explain what happens when $\omega \to 0$.

Solution:

(a) $\frac{dI}{dt} + \frac{R}{L}I = \frac{V_0}{L}\cos(\omega t)$. Let $\alpha = R/L$.

IF: $e^{\alpha t}$. $\frac{d}{dt}(Ie^{\alpha t}) = \frac{V_0}{L}e^{\alpha t}\cos(\omega t)$.

The PI requires $I_p = A\cos(\omega t) + B\sin(\omega t)$. $I_p' = -A\omega\sin(\omega t) + B\omega\cos(\omega t)$. $\alpha I_p + I_p' = \alpha A\cos + \alpha B\sin - A\omega\sin + B\omega\cos = \frac{V_0}{L}\cos(\omega t)$.

$\alpha A + B\omega = V_0/L$ and $\alpha B - A\omega = 0$So $B = A\omega/\alpha$.

$A\alpha + A\omega^2/\alpha = V_0/L$. $A = \frac{V_0\alpha}{L(\alpha^2 + \omega^2)} = \frac{V_0 R}{L^2(\omega^2 + R^2/L^2)} = \frac{V_0 R}{R^2 + \omega^2 L^2}$.

$I = Ce^{-Rt/L} + \frac{V_0}{R^2 + \omega^2 L^2}[R\cos(\omega t) + \omega L\sin(\omega t)]$.

At $t = 0$: $0 = C + \frac{V_0 R}{R^2 + \omega^2 L^2}$So $C = -\frac{V_0 R}{R^2 + \omega^2 L^2}$.

(b) Transient: $Ce^{-Rt/L}$ (decays to zero as $t \to \infty$). Steady-state: $\frac{V_0}{R^2 + \omega^2 L^2}[R\cos(\omega t) + \omega L\sin(\omega t)]$.

(c) Amplitude $= \frac{V_0}{R^2 + \omega^2 L^2}\sqrt{R^2 + \omega^2 L^2} = \frac{V_0}{\sqrt{R^2 + \omega^2 L^2}}$.

(d) When $\omega \to 0$ (DC circuit): amplitude $\to \frac{V_0}{R} = I_0$. The current approaches $V_0/R$ with time constant $L/R$. This is the classic RL circuit charging: the inductor initially opposes current flow, but as the magnetic field builds, the current asymptotically approaches $V_0/R$.

Forgetting the constant of integration: When solving differential equations, always include the constant $C$ (or use initial conditions to find it). A common error is to solve the homogeneous part and particular integral but forget to combine them with the arbitrary constant, giving an incomplete general solution.

Confusing the homogeneous and particular solutions: The homogeneous solution solves $\frac{dy}{dx} + Py = 0$ and gives the transient behaviour (decays to zero). The particular solution solves the full equation with the forcing term and gives the steady-state behaviour. The general solution is the sum of both. Students often write only one and forget the other.

Mixing up the integrating factor method: For equations of the form $\frac{dy}{dx} + Py = Q$, the integrating factor is $\mu = e^{\int P\,dx}$. Multiply both sides by $\mu$, then recognise the left side as $\frac{d}{dx}(\mu y)$. A common error is to forget to multiply the right-hand side $Q$ by the integrating factor.

• Differential Equations: Differential equations require calculus techniques
• Further Calculus: Calculus underpins further mathematics
• Pure Mathematics: Further maths extends A-level mathematics