This article is aimed at undergraduate students who are struggling with differential equations, or anyone interested in self-study.

A differential equation is a mathematical equation that describes the relationship between a function's derivatives and the function itself, as well as other variables.

Classification of Differential Equations

To master the computation of differential equations, one must first learn to classify differential equations in order to choose the correct method for solving them.

Partial differential equations and higher-order ordinary differential equations are not covered in this article, which focuses on ordinary differential equations that can be roughly classified into four categories:

Separable Variable Equations

• Form: Can be expressed in the form $N(y)dy=M(x)dx$, where $N(y)$ and M(x) are functions of y and x.
• Characteristic: The equation can be rearranged so that all terms involving y are on one side and all terms involving x are on the other side.

First-Order Linear Homogeneous Differential Equations

• Form: $y^{\prime}+ P(x)y=0$

Second-Order Linear Homogeneous Differential Equations

• Form: $y^{\prime\prime}+py^{\prime}+qy=0$.
• Characteristic: The equation is linear and has no constant term (i.e., no "non-homogeneous" term). The coefficients of $y$ and its derivatives are functions of x only.

Second-Order Linear Non-Homogeneous Differential Equations

• Form: $y^{\prime\prime}+py^{\prime}+qy=Q(x)$
• Characteristic: In addition to being linear, it includes a non-homogeneous term Q(x) that is not equal to zero.

Bernoulli's Equation

• Form: d**y/d**x+P(x)y=Q(x)y**n, where n is not equal to 0 or 1.
• Characteristic: This is a nonlinear equation. It is characterized by the presence of a term involving $y^n$, where n is a real number.

Solution Methods

Separable Variable Equations

Move all terms involving y to one side of the equation and all terms involving $x$ to the other side, then integrate both sides.

First-Order Homogeneous Differential Equations

Simply remember the general solution formula:

$y = Ce^{-\int P(x)\mathrm{d}x}$

First-Order Non-Homogeneous Differential Equations

Simply remember the general solution formula:

$y = Ce^{-\int P(x)\mathrm{d}x} + e^{-\int P(x)\mathrm{d}x}\int Q(x)e^{\int P(x) \mathrm{d}x}\mathrm{d}x$

Memory tip: General solution of the homogeneous form + particular solution. The term $e^{-\int P(x) \mathrm{d}x}$ is the integrating factor.

Second-Order Homogeneous Differential Equations

For this type of equation, there are two methods of solving.

We will explain using this equation as an example:

$y^{\prime\prime} - 4y^{\prime} + 13y = 0$

One method is the characteristic equation method. The steps are as follows:

1. Write the characteristic equation: $r^2 - 4r + 13 = 0$

2. Solve the characteristic equation: Calculate it as a quadratic equation.

3. Write the general solution:

4. If the characteristic equation has two distinct roots, the general solution is $y = C\_1e^{r\_1x} + C\_2e^{r\_2x}$

5. If the roots of the characteristic equation are repeated (i.e., $r\_1 = r\_2$), the general solution is $e^{rx}(C\_1+C\_2x)$

6. If the roots of the characteristic equation are complex conjugates, the general solution is $e^{ax}(C\_1\cos(bx) + C\_2\sin(bx))$

You may wonder, what are complex conjugates? They refer to the case where the discriminant of the characteristic equation $\Delta < 0$. Generally, we would say that this equation has no real solutions because the square root cannot be negative.

In this case, if we take the absolute value under the square root and then add the imaginary unit $i$ to the result after taking the square root, we can still obtain two roots. For example, for the characteristic equation:

$r^2 + 4r + 13 = 0$

Using the quadratic formula, we find the roots to be:

$\frac{-4 + \sqrt{16 - 62}}{2}$

Here, the expression under the square root is -36, leading us to the complex conjugate roots:

$r\_1 = \frac{-4+6i}{2} = -2 + 3i\\\ r\_2 = \frac{-4-6i}{2} = -2 - 3i$

Where $a=-2, b=3$.

Thus, the general solution for this equation is:

$y = e^{-2x}(C\_1\sin(3x)+C\_2\cos(3x))$

Second-Order Non-Homogeneous Differential Equations

For this type of equation, we first assume $Q(x)$ to be 0 to compute the general solution $y\_h$.

Next, we need to find a particular solution $y\_p$.

Based on the form of $Q(x)$, we set a trial particular solution. For example, consider this equation:

$y^{\prime\prime} - 3y^{\prime} + 2 = e^x$

Assume the trial solution to be $y = Axe^x$, substituting it into the original equation:

$(Axe^x)^{\prime\prime} - 3(Axe^x)^{\prime} + 2 = e^x$

After simplification, we get:

$(2Ae^x + Axe^x) + 3(Ae^x + Axe^x) +2 = e^x$

Solving gives:

$A = -1$

Next, substituting the trial solution back into the original equation, the solution to the non-homogeneous differential equation is the general solution of the homogeneous equation plus the particular solution of the non-homogeneous equation:

$y = y\_{h} + y\_{p}$

Thus, the general solution of the original equation is:

$y = C\_1e^x + C\_2e^{2x} - e^x$

For determining the trial particular solution, common forms include:

• $\cos(wt)$ or $\sin(wt)$: $a\cos(wt) + b\sin(wt)$
• $e^{at}$: $Be^{at}$

Example Problem

Given that the function f(x) is a solution to the differential equation $y^{\prime\prime} - 2y^{\prime} + 5y = 4e^x$ and that $f(0) = -2$, $f^{\prime}(0) = -2$, find $f(x)$.

Summary

The difficulty in solving differential equations lies in memorizing the general solution formulas, but overall, it is not too challenging.