FE Mathematics: Differential Equations Review

Differential equations are the mathematical engine behind nearly every dynamic system in engineering, from circuit analysis to heat transfer and mechanical vibrations. On the FE exam, you won’t have time to derive every solution from scratch; your success hinges on quickly recognizing the type of equation and selecting the most efficient, reliable solution path. This review distills the core ODE methods you must master, focusing on strategic problem-solving for exam efficiency.

First-Order Ordinary Differential Equations: Recognition and Resolution

First-order ODEs form the foundational layer tested on the exam. They are typically presented in the form $dy/dx=f(x,y)$ or equivalently $M(x,y)dx+N(x,y)dy=0$. Your first job is to classify the equation to unlock the correct solution method.

Separable equations are the simplest type, where you can algebraically manipulate the equation to isolate all $x$ terms with $dx$ and all $y$ terms with $dy$. The general form is $g(y)dy=h(x)dx$. The solution is obtained by direct integration: $\int g(y)dy=\int h(x)dx+C$. For example, to solve $dy/dx=x{y}^2$, you separate: $y^{-2}dy=xdx$, then integrate both sides.

Linear first-order equations follow the standard form $dy/dx+P(x)y=Q(x)$. Here, the key is the integrating factor, $\mu (x)=e^{\int P(x)dx}$. You multiply the entire equation by this factor, which cleverly makes the left side a perfect derivative: $d/dx[\mu (x)y]=\mu (x)Q(x)$. You then integrate with respect to $x$: $\mu (x)y=\int \mu (x)Q(x)dx+C$, and solve for $y$. This method is systematic and works every time the equation is linear.

Exact equations are a specific class where a function $F(x,y)$ exists such that its total differential $dF=Mdx+Ndy$ equals zero. The test for exactness is: $\partial M/\partial y=\partial N/\partial x$. If this condition holds, you find $F$ by integrating: $F(x,y)=\int M(x,y)dx+h(y)$, where $h(y)$ is a function of $y$ found by ensuring $\partial F/\partial y=N(x,y)$. The solution is then $F(x,y)=C$. If the equation is not exact, the exam might hint at an integrating factor to make it exact, but this is a less common twist.

Second-Order Linear ODEs with Constant Coefficients

These equations model oscillatory and decay/growth phenomena and are paramount for the FE. The general form is $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ for the homogeneous case, and $=g(x)$ for the non-homogeneous case. The solution is $y=y_h+y_p$, where $y_h$ is the complementary solution from the homogeneous equation and $y_p$ is a particular solution.

To find $y_h$, you solve the characteristic equation: $a{r}^2+br+c=0$. The roots $r_1$ and $r_2$ dictate the form:

• Real and distinct roots: $y_h=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$
• Real and equal roots (repeated): $y_h=C_1{e}^{rx}+C_{2}x{e}^{rx}$
• Complex conjugate roots $r=\alpha \pm \beta i$: $y_h=e^{\alpha x}(C_1\cos (\beta x)+C_2\sin (\beta x))$

Memorizing these three solution forms saves critical time. You will almost always need to apply initial conditions (like $y(0)$ and $y^{\prime }(0)$) to solve for the constants $C_1$ and $C_2$ later.

Finding Particular Solutions: The Method of Undetermined Coefficients

When the non-homogeneous term $g(x)$ is a simple function (like polynomials, exponentials, sines/cosines, or sums/products thereof), the method of undetermined coefficients is your fastest tool. You postulate a "guess" for $y_p$ based on the form of $g(x)$, but with undetermined coefficients (A, B, etc.).

For example, if $g(x)=5e^{2x}$, guess $y_p=A{e}^{2x}$. If $g(x)=3x^2$, guess $y_p=A{x}^2+Bx+C$. The crucial exception is if your guess is part of the homogeneous solution $y_h$; in that case, you must multiply your guess by $x$ (or $x^2$ for a double root) to ensure linear independence. After making the correct guess, you substitute $y_p$ and its derivatives back into the original ODE and equate coefficients to solve for A, B, etc.

The Power of the Laplace Transform Method

For linear ODEs with constant coefficients, especially those with discontinuous or impulsive forcing functions (like a sudden load or switched voltage), the Laplace transform is immensely powerful. It converts a differential equation in the time domain (t) into an algebraic equation in the s-domain. The key operational transforms to know are: $$\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)$$ $$\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)$$

The solution process is methodical:

1. Take the Laplace transform of every term in the ODE, using the initial conditions.
2. Solve the resulting algebraic equation for $Y(s)$.
3. Perform the inverse Laplace transform on $Y(s)$ to get $y(t)$, the solution in the time domain. This often involves partial fraction decomposition to break $Y(s)$ into simpler, recognizable transforms.

This method elegantly handles initial value problems in one consolidated step and is often the best choice for equations with terms like $u_c(t)$ (unit step functions).

Common Pitfalls

1. Misapplying the undetermined coefficients guess: The most common error is forgetting to modify your guess when it overlaps with the homogeneous solution. If $g(x)=e^{3x}$ and $e^{3x}$ is already in $y_h$, your guess must be $Ax{e}^{3x}$, not $A{e}^{3x}$. Always compare your guess to $y_h$ first.

1. Incorrect characteristic equation for second-order ODEs: For the equation $2y^{\prime \prime }+3y^{\prime }-5y=0$, the characteristic equation is $2r^2+3r-5=0$. A frequent slip is writing the coefficients incorrectly (e.g., $r^2+3r-5=0$). Ensure the coefficients $a$, $b$, and $c$ match the ODE exactly.

1. Forgetting the product rule in exact equations: When integrating $M(x,y)$ with respect to $x$ to find $F(x,y)$, remember your "constant" of integration is actually a function $h(y)$. You then find $h(y)$ by taking $\partial F/\partial y$ and setting it equal to $N(x,y)$. Neglecting this step leaves an incomplete solution.

1. Laplace transform domain confusion: Remember that after transforming, you are solving for $Y(s)$. A common algebraic error is to treat $s$ as a time-domain variable. Stay organized and carefully perform the algebra in the s-domain before looking up the inverse transform.

Summary

• Classify first-order ODEs immediately: Use separation of variables for $g(y)dy=h(x)dx$, the integrating factor for linear equations $y^{\prime }+P(x)y=Q(x)$, and check for exactness if in the form $Mdx+Ndy=0$.
• Solve second-order linear homogeneous ODEs via the characteristic equation $a{r}^2+br+c=0$, and memorize the three solution forms based on root types (real distinct, real repeated, complex conjugate).
• For non-homogeneous equations, use undetermined coefficients for simple $g(x)$ functions, but always modify your guess by $x$ or $x^2$ if it duplicates part of the homogeneous solution.
• The Laplace transform is the tool of choice for IVPs with discontinuous forcing; it incorporates initial conditions upfront and turns differentiation into algebra.
• Your primary exam skill is pattern recognition. Quickly identify the equation type to select the fastest, most appropriate solution path and avoid time-consuming algebraic detours.