Key Concepts & Formulas
- $$\text{Separable: } \frac{dy}{dx} = f(x)g(y)$$
- $$\mu(x) = e^{\int P(x)\,dx}$$
- $$ar^2 + br + c = 0$$
- Variation of parameters
- $$\mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t)\,dt$$
- $$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$
- $$\text{Convolution: } (f*g)(t) = \int_0^{t} f(\tau)g(t-\tau)\,d\tau$$
- $$W(y_1, y_2) = y_1 y_2' - y_1' y_2$$
- Sturm-Liouville theory
- Phase plane analysis for systems
1Easy
Solve the separable ODE: \(\frac{dy}{dx} = \frac{x^2 + 1}{y^2}\), with \(y(0) = 1\).
Show Hint
Separate: \(y^2\,dy = (x^2 + 1)\,dx\). Integrate both sides: \(\frac{y^3}{3} = \frac{x^3}{3} + x + C\). Apply the initial condition to find \(C = \frac{1}{3}\).
2Medium
Solve the first-order linear ODE: \(y' + \frac{2}{x}y = x^2\), for \(x > 0\).
Show Hint
The integrating factor is \(\mu(x) = e^{\int (2/x)\,dx} = x^2\). Multiply through: \((x^2 y)' = x^4\). Integrate: \(x^2 y = \frac{x^5}{5} + C\), so \(y = \frac{x^3}{5} + \frac{C}{x^2}\).
3Medium
Find the general solution of \(y'' - 4y' + 13y = 0\). Then solve the IVP with \(y(0) = 1\), \(y'(0) = 6\).
Show Hint
Characteristic equation: \(r^2 - 4r + 13 = 0\), giving \(r = 2 \pm 3i\). General solution: \(y = e^{2x}(c_1 \cos 3x + c_2 \sin 3x)\). Apply initial conditions to find \(c_1 = 1\), then differentiate and use \(y'(0) = 6\).
4Hard
Solve \(y'' + y = \sec(x)\) using variation of parameters.
Show Hint
The complementary solution is \(y_c = c_1 \cos(x) + c_2 \sin(x)\). For variation of parameters: \(u_1' = -\sin(x)\sec(x) = -\tan(x)\), \(u_2' = \cos(x)\sec(x) = 1\). Integrate to get \(u_1 = \ln|\cos(x)|\) and \(u_2 = x\).
5Medium
Use Laplace transforms to solve: \(y'' + 4y = \delta(t - \pi)\), \(y(0) = 0\), \(y'(0) = 0\), where \(\delta\) is the Dirac delta function.
Show Hint
Take the Laplace transform: \(s^2 Y + 4Y = e^{-\pi s}\). So \(Y(s) = \frac{e^{-\pi s}}{s^2 + 4}\). Inverse transform using the time-shift: \(y(t) = \frac{1}{2}\sin(2(t - \pi))u(t - \pi) = -\frac{1}{2}\sin(2t)u(t - \pi)\).
6Hard
Solve the Bernoulli equation: \(y' + y = xy^3\). Find both the general solution and any singular solutions.
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Divide by \(y^3\): \(\frac{y'}{y^3} + y^{-2} = x\). Substitute \(v = y^{-2}\), so \(v' = -2y'/y^3\). This gives \(v' - 2v = -2x\), a first-order linear ODE. Also check \(y = 0\) as a singular solution.
7Hard
Find the general solution of the system: \(x' = 3x - y\), \(y' = 4x - y\). Classify the critical point at the origin and sketch the phase portrait.
Show Hint
The coefficient matrix \(A = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}\) has eigenvalues \(\lambda = 1\) (repeated). Check: \((A - I)\) is not zero, so there's a defective eigenvalue. Use a generalized eigenvector. The origin is an unstable improper node.
8Easy
Verify that \(y_1 = e^x\) and \(y_2 = xe^x\) are solutions of \(y'' - 2y' + y = 0\). Compute their Wronskian and verify linear independence.
Show Hint
Substitute each into the ODE. For the Wronskian: \(W = y_1 y_2' - y_1' y_2 = e^x(e^x + xe^x) - e^x(xe^x) = e^{2x} \neq 0\), confirming linear independence.
9Medium
Find the convolution \(f * g\) where \(f(t) = e^{2t}\) and \(g(t) = e^{3t}\). Verify your answer using Laplace transforms.
Show Hint
\((f*g)(t) = \int_0^{t} e^{2\tau} e^{3(t-\tau)}\,d\tau = e^{3t}\int_0^{t} e^{-\tau}\,d\tau = e^{3t}(1 - e^{-t}) = e^{3t} - e^{2t}\). Check: \(\mathcal{L}\{f*g\} = F(s)G(s) = \frac{1}{(s-2)(s-3)}\).
10Hard
Solve the boundary value problem: \(y'' + \lambda y = 0\), \(y(0) = 0\), \(y(L) = 0\). Find all eigenvalues \(\lambda\) and corresponding eigenfunctions.
Show Hint
Consider three cases: \(\lambda < 0\), \(\lambda = 0\), \(\lambda > 0\). Only \(\lambda > 0\) gives nontrivial solutions. Setting \(\lambda = \left(\frac{n\pi}{L}\right)^2\) for \(n = 1, 2, 3, \ldots\), the eigenfunctions are \(y_n(x) = \sin\!\left(\frac{n\pi x}{L}\right)\).
11Medium
Solve the exact ODE: \((2xy + 3)\,dx + (x^2 + 4y)\,dy = 0\). Verify exactness first, then find the solution.
Show Hint
Check: \(\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}\). So it is exact. Find \(F\) such that \(F_x = 2xy + 3\) and \(F_y = x^2 + 4y\). Integrating: \(F = x^2 y + 3x + 2y^2 = C\).
12Hard
Consider the nonlinear system \(x' = x - xy\), \(y' = -y + xy\) (Lotka-Volterra predator-prey). Find all equilibria, linearize at each, and classify their stability.
Show Hint
Equilibria: \((0,0)\) and \((1,1)\). At \((0,0)\), the Jacobian is \(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) — a saddle point. At \((1,1)\), the Jacobian is \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) with eigenvalues \(\pm i\) — a center (in the linearization). The nonlinear system has a conserved quantity, confirming periodic orbits.