Common Mistakes to Avoid | Math 4

Ch 1 First-Order ODEs ▲

1 Forgetting the constant of integration after separating

Wrong

\(\displaystyle \int \frac{dy}{y} = \int x\,dx \implies \ln|y| = \frac{x^2}{2}\)

Correct

\(\displaystyle \ln|y| = \frac{x^2}{2} + C\)

Tip: Always add \(+C\) immediately when you integrate. Missing it gives a single curve instead of the full family of solutions.

2 Dividing by \(y\) without checking \(y=0\)

Wrong

\(\displaystyle y' = xy^2 \implies \frac{y'}{y^2} = x\) (no comment about \(y=0\))

Correct

Note that \(y=0\) is a solution. For \(y\neq 0\), divide by \(y^2\) and proceed.

Tip: Whenever you divide by an expression involving \(y\), check whether that expression being zero gives a valid (possibly singular) solution.

3 Forgetting the integrating factor in linear equations

Wrong

\(y' + 2y = e^x \implies\) integrate both sides directly: \(y = e^x + C\)

Correct

Multiply by \(\mu = e^{\int 2\,dx} = e^{2x}\). Then \((e^{2x}y)' = e^{3x}\), so \(y = \frac{1}{3}e^x + Ce^{-2x}\).

Tip: A linear ODE \(y'+P(x)y=Q(x)\) is never solved by "just integrating." You must use the integrating factor \(\mu = e^{\int P\,dx}\).

4 Wrong substitution in Bernoulli equations

Wrong

\(y' + y = y^3\) — substituting \(v = y\) or \(v = y^3\)

Correct

For \(y' + Py = Qy^n\) with \(n=3\): use \(v = y^{1-n} = y^{-2}\).

Tip: The Bernoulli substitution is always \(v = y^{1-n}\). Memorize the formula, don't guess.

5 Confusing exact equations with separable ones

Wrong

\((2xy + 3)\,dx + (x^2 + 4y)\,dy = 0\) — trying to separate \(x\) and \(y\)

Correct

Check: \(\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}\). It's exact — find \(F(x,y)\) such that \(F_x = M,\; F_y = N\).

Tip: Before choosing a method, classify the equation. Check separable → linear → exact → Bernoulli in order.

6 Dropping the absolute value in \(\ln|y|\)

Wrong

\(\ln y = x + C \implies y = e^{x+C}\) (only positive \(y\))

Correct

\(\ln|y| = x + C \implies y = \pm e^C \cdot e^x = Ae^x\) where \(A\) can be positive or negative.

Ch 2 Higher-Order Linear ODEs ▲

1 Wrong form of \(y_p\) when it overlaps with \(y_h\)

Wrong

\(y'' - 3y' + 2y = e^{2x}\). Guessing \(y_p = Ae^{2x}\) — but \(e^{2x}\) is already in \(y_h\)!

Correct

Since \(e^{2x}\) solves the homogeneous equation, multiply by \(x\): try \(y_p = Axe^{2x}\).

Tip: Always find \(y_h\) first. If your \(y_p\) guess duplicates any term in \(y_h\), multiply by \(x\) (or \(x^2\) if needed) until no overlap.

2 Forgetting to include all terms in the particular guess

Wrong

For \(g(x) = 3x^2\), guessing \(y_p = Ax^2\) only

Correct

Guess \(y_p = Ax^2 + Bx + C\) — include all lower-degree terms.

Tip: In undetermined coefficients, if \(g(x)\) contains \(x^n\), your guess must include \(x^n, x^{n-1}, \ldots, x^0\).

3 Writing the wrong characteristic equation

Wrong

\(y'' + 5y' + 6y = 0 \implies r^2 + 5r + 6y = 0\) (leftover \(y\))

Correct

Replace \(y''\to r^2\), \(y'\to r\), \(y\to 1\): \(r^2 + 5r + 6 = 0\).

4 Mishandling complex roots

Wrong

Roots \(r = 2 \pm 3i\) — writing \(y = c_1 e^{(2+3i)x} + c_2 e^{(2-3i)x}\) as the final answer

Correct

\(y = e^{2x}(c_1\cos 3x + c_2\sin 3x)\)

Tip: For roots \(\alpha \pm \beta i\), always convert to the real form \(e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)\).

5 Using undetermined coefficients when the ODE has variable coefficients

Wrong

\(x^2 y'' + xy' - y = x^3\) — trying to guess \(y_p = Ax^3\)

Correct

Undetermined coefficients only works for constant-coefficient equations. Use variation of parameters instead.

6 Euler-Cauchy: substituting \(y=e^{rx}\) instead of \(y=x^r\)

Wrong

\(x^2y'' + 3xy' + y = 0 \implies\) try \(y = e^{rx}\)

Correct

For Euler-Cauchy equations \(ax^2y'' + bxy' + cy = 0\), use \(y = x^r\).

Ch 3 Laplace Transforms ▲

1 Forgetting initial conditions in the transform of \(y''\)

Wrong

\(\mathcal{L}\{y''\} = s^2 Y(s)\)

Correct

\(\mathcal{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)\)

Tip: The whole point of Laplace transforms is to absorb initial conditions. \(\mathcal{L}\{y'\} = sY - y(0)\) and \(\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)\). Never drop these terms.

2 Wrong partial-fraction setup

Wrong

\(\frac{1}{(s-1)(s^2+1)} = \frac{A}{s-1} + \frac{B}{s^2+1}\)

Correct

\(\frac{1}{(s-1)(s^2+1)} = \frac{A}{s-1} + \frac{Bs + C}{s^2+1}\)

Tip: An irreducible quadratic \(s^2+as+b\) in the denominator needs \(\frac{Bs+C}{s^2+as+b}\), not just a single constant.

3 Confusing \(\mathcal{L}\{e^{at}f(t)\}\) with \(e^{as}F(s)\)

Wrong

\(\mathcal{L}\{e^{2t}\sin t\} = e^{2s}\cdot\frac{1}{s^2+1}\)

Correct

First shift theorem: \(\mathcal{L}\{e^{at}f(t)\} = F(s-a)\). So \(\mathcal{L}\{e^{2t}\sin t\} = \frac{1}{(s-2)^2+1}\).

Tip: \(e^{at}\) in the time domain shifts \(s\) → \(s-a\). The factor \(e^{-as}\) in the s-domain is a time delay (second shift theorem). Don't mix them up.

4 Ignoring the unit step in the second shift theorem

Wrong

\(\mathcal{L}^{-1}\{e^{-3s}\cdot\frac{1}{s^2}\} = (t-3)^2\) (missing step function, wrong power)

Correct

\(\mathcal{L}^{-1}\!\left\{\frac{1}{s^2}\right\} = t\), so \(\mathcal{L}^{-1}\!\left\{e^{-3s}\cdot\frac{1}{s^2}\right\} = (t-3)\,u(t-3)\).

5 Applying Laplace to a non-IVP without adjusting

Wrong

Using Laplace on a boundary value problem \(y(0)=1, y(\pi)=0\) as if both conditions are at \(t=0\)

Correct

Laplace transforms require initial conditions at \(t=0\). For BVPs, use other methods (eigenvalue, shooting, etc.).

Ch 4 Systems of ODEs ▲

1 Using wrong eigenvectors (row-reducing incorrectly)

Wrong

For \(A = \begin{pmatrix}2 & 1\\0 & 3\end{pmatrix}\), eigenvalue \(\lambda=2\): solving \((A-2I)\mathbf{v}=0\) and picking \(\mathbf{v}=\begin{pmatrix}1\\1\end{pmatrix}\)

Correct

\(A-2I = \begin{pmatrix}0 & 1\\0 & 1\end{pmatrix}\), giving \(v_2=0\). So \(\mathbf{v}=\begin{pmatrix}1\\0\end{pmatrix}\).

Tip: After finding \(\lambda\), always verify by checking \(A\mathbf{v} = \lambda\mathbf{v}\).

2 Forgetting the \(te^{\lambda t}\) term for repeated eigenvalues

Wrong

Repeated \(\lambda\): writing \(\mathbf{x} = c_1\mathbf{v}e^{\lambda t} + c_2\mathbf{v}e^{\lambda t}\) (same thing twice)

Correct

\(\mathbf{x} = c_1\mathbf{v}e^{\lambda t} + c_2(\mathbf{v}t + \mathbf{w})e^{\lambda t}\) where \((A-\lambda I)\mathbf{w} = \mathbf{v}\).

3 Misreading phase portrait stability

Wrong

Eigenvalues \(\lambda = -1, 3\) — calling it a "stable node"

Correct

One positive, one negative → saddle point (unstable). Both must be negative for a stable node.

Tip: Stable node: both \(\lambda < 0\). Unstable node: both \(\lambda > 0\). Saddle: opposite signs. Spiral: complex \(\lambda\).

4 Writing complex eigenvector solutions without converting to real form

Wrong

Leaving the answer as \(\mathbf{x} = c_1\begin{pmatrix}1\\i\end{pmatrix}e^{(2+3i)t} + c_2\begin{pmatrix}1\\-i\end{pmatrix}e^{(2-3i)t}\)

Correct

Decompose into real and imaginary parts to get solutions involving \(e^{2t}\cos 3t\) and \(e^{2t}\sin 3t\) with real vectors.

Ch 5 Series & Numerical Methods ▲

1 Wrong index shift when substituting power series

Wrong

\(y = \sum_{n=0}^{\infty} a_n x^n \implies y'' = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}\) — starting at \(n=0\) without re-indexing

Correct

\(y'' = \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}\). To align powers, let \(m=n-2\): \(\sum_{m=0}^{\infty}(m+2)(m+1)a_{m+2}x^m\).

Tip: When combining series, every sum must have the same power of \(x\) and the same starting index. Re-index carefully.

2 Using ordinary power series at a singular point

Wrong

\(x^2y'' + xy' + y = 0\) — assuming \(y = \sum a_n x^n\) around \(x=0\)

Correct

\(x=0\) is a regular singular point. Use the Frobenius method: \(y = \sum a_n x^{n+r}\).

Tip: Always check: is the point ordinary or singular? If \(P(x_0)=0\) in the standard form, you need Frobenius.

3 Using a step size that is too large in Euler's method

Wrong

\(h = 1.0\) for \(y' = -10y\) — solution blows up (instability)

Correct

Use \(h < 0.2\) (stability requires \(|1 + h\lambda| < 1\)). Smaller \(h\) = more stable, more accurate.

Tip: For stiff equations (large \(|\lambda|\)), Euler's method needs very small \(h\). Runge-Kutta 4 is much more forgiving.

4 Confusing the indicial equation with the recurrence relation

Wrong

Setting the \(n=0\) coefficient equation to find \(a_1\) instead of \(r\)

Correct

The lowest-power coefficient gives the indicial equation for \(r\). Only after finding \(r\) do you use higher-power equations as the recurrence for \(a_n\).

Ch 6 Partial Differential Equations ▲

1 Applying separation of variables to non-homogeneous BCs directly

Wrong

\(u(0,t) = 0,\; u(L,t) = T_1 \neq 0\) — directly writing \(u(x,t) = X(x)T(t)\)

Correct

First find the steady-state \(u_s(x)\). Then let \(v = u - u_s\), which satisfies homogeneous BCs. Apply separation to \(v\).

Tip: Separation of variables only works with homogeneous boundary conditions. Always subtract the steady-state first.

2 Dismissing \(\lambda = 0\) in the eigenvalue analysis without checking

Wrong

"Case 1: \(\lambda = 0\) gives trivial solution" — always, for any BC type

Correct

For Neumann BCs \((X'(0)=0, X'(L)=0)\), the case \(\lambda=0\) gives \(X(x) = 1\) (a constant), which is non-trivial. This is the DC mode.

Tip: Always work through all three cases (\(\lambda=0\), \(\lambda<0\), \(\lambda>0\)) for every BC type. What's trivial for Dirichlet may be non-trivial for Neumann.

3 Using \(\sin\) eigenfunctions with Neumann BCs (or \(\cos\) with Dirichlet)

Wrong

Dirichlet BCs \(u(0,t)=0, u(L,t)=0\) — writing \(X_n = \cos\frac{n\pi x}{L}\)

Correct

Dirichlet \((X(0)=X(L)=0)\) → \(\sin\). Neumann \((X'(0)=X'(L)=0)\) → \(\cos\). Each BC type selects its own eigenfunction family.

4 Confusing the heat equation solution structure with the wave equation

Wrong

Wave equation: writing \(T_n(t) = e^{-\alpha n^2 \pi^2 t/L^2}\) (exponential decay)

Correct

Heat: \(T_n(t) = e^{-k\lambda_n t}\) (decay). Wave: \(T_n(t) = A_n\cos(c\lambda_n t) + B_n\sin(c\lambda_n t)\) (oscillation).

Tip: Heat dissipates (exponential decay). Waves oscillate (sin/cos in time). The spatial eigenfunctions are the same; only the temporal part differs.

5 Computing Fourier coefficients with the wrong integral

Wrong

\(B_n = \int_0^L f(x)\sin\frac{n\pi x}{L}\,dx\) (missing the \(\frac{2}{L}\))

Correct

\(B_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx\)

Tip: The \(\frac{2}{L}\) factor comes from the orthogonality integral \(\int_0^L \sin^2\frac{n\pi x}{L}\,dx = \frac{L}{2}\). Never forget it.

Ch 7 Introduction to Complex Numbers ▲

1 Using \(\arctan(b/a)\) for the argument without checking the quadrant

Wrong

For \(z = -1 + i\): \(\theta = \arctan(1/(-1)) = -\pi/4\)

Correct

\(z=-1+i\) is in the second quadrant, so \(\theta = \pi - \pi/4 = 3\pi/4\)

Tip: \(\arctan\) only returns angles in \((-\pi/2, \pi/2)\), so it can never land in quadrant II or III. Sketch the point first, or use atan2(b, a).

2 Forgetting that \(i^2 = -1\) changes the sign when multiplying

Wrong

\((2+3i)(1+4i) = 2 + 8i + 3i + 12i^2 = 2 + 11i + 12i^2\), then leaving \(12i^2\) as is

Correct

\(12i^2 = -12\), so the result is \((2-12) + 11i = -10 + 11i\)

Tip: After every multiplication, scan for \(i^2\) and replace it with \(-1\) before collecting terms.

3 Dividing without multiplying by the conjugate

Wrong

\(\dfrac{3+i}{2-i}\) — splitting into \(\dfrac{3}{2} + \dfrac{i}{-i}\)

Correct

Multiply top and bottom by \(2+i\): \(\dfrac{(3+i)(2+i)}{(2-i)(2+i)} = \dfrac{5+5i}{5} = 1+i\)

Tip: A complex number is not in standard form until the denominator is real. The conjugate is the only reliable way to get there.

4 Reporting only one nth root

Wrong

“The cube root of \(8i\) is \(2e^{i\pi/6}\).”

Correct

There are three: \(2e^{i\pi/6},\; 2e^{i5\pi/6},\; 2e^{i3\pi/2}\)

Tip: An nth root always has exactly \(n\) answers, equally spaced by \(2\pi/n\) on a circle. Let \(k = 0, 1, \ldots, n-1\).

5 Leaving the DE solution in complex exponential form

Wrong

Roots \(-2 \pm 3i\), so \(y = c_1e^{(-2+3i)t} + c_2e^{(-2-3i)t}\) — and stopping there

Correct

Convert to the real form \(y = e^{-2t}(c_1\cos 3t + c_2\sin 3t)\)

Tip: Physical problems need real-valued answers. Use Euler's formula to convert; the real part becomes the decay envelope and the imaginary part becomes the frequency.

6 Swapping the roles of \(\alpha\) and \(\beta\) in \(\alpha \pm \beta i\)

Wrong

Roots \(-1 \pm 4i\) giving \(y = e^{4t}(c_1\cos t + c_2\sin t)\)

Correct

\(y = e^{-t}(c_1\cos 4t + c_2\sin 4t)\) — real part in the exponential, imaginary part in the trig

Tip: Remember the physics: the real part controls how fast it dies away, the imaginary part controls how fast it wiggles.