A differential equation gives you a derivative and asks you to find the function behind it. When I first worked through these problems, integration was rarely the hardest part. Choosing the right method was.
These solved examples of differential equations show three common first-order methods: separation of variables, an integrating factor, and substitution with y = vx. Every equation below uses plain text so you can copy it directly into Google Docs.
How Do You Choose the Right Method?
I start by looking at the equation’s shape. If I can put every y term with dy and every x term with dx, I separate the variables. If the equation has the form dy/dx + P(x)y = Q(x), I check for an integrating factor. If its right side depends only on the ratio y/x, I try y = vx.
An equation can fit more than one description. The third example, for instance, is both linear and homogeneous in the substitution sense. I use it to demonstrate the y = vx method because that pattern appears often in introductory exercises.
MIT OpenCourseWare’s differential equations materials provide further lessons on these first-order methods.
Example 1: Solve a Separable Differential Equation
Find the general solution to:
dy/dx = x^2/y
Separate and integrate
Multiply by y dx to put each variable on its own side:
y dy = x^2 dx
Integrate both sides:
∫ y dy = ∫ x^2 dx
y^2/2 = x^3/3 + C1
Multiply by 2. Since 2C1 is still an arbitrary constant, call it C:
y^2 = 2x^3/3 + C
That implicit equation is a useful final answer. If you need y explicitly, take the square root:
y = ±sqrt(2x^3/3 + C)
The expression inside the square root must be positive wherever the original equation applies. Also, y cannot equal zero because the original right side contains 1/y.
Check the solution
Differentiate the implicit answer:
y^2 = 2x^3/3 + C
2y(dy/dx) = 2x^2
dy/dx = x^2/y
The result matches the given equation. I prefer this check because it verifies both square-root branches at once.
If you want more practice recognizing when separation works, OpenStax’s separable equations section develops the method with additional examples.
Example 2: Solve a First-Order Linear Differential Equation
Find the general solution to:
dy/dx + 2y = e^x
This matches dy/dx + P(x)y = Q(x), with P(x) = 2 and Q(x) = e^x. The integrating factor turns the left side into one product derivative.
Find and apply the integrating factor
Calculate the factor:
IF = e^(∫P(x) dx)
= e^(∫2 dx)
= e^(2x)
Multiply every term in the original equation by e^(2x):
e^(2x)(dy/dx) + 2e^(2x)y = e^(3x)
The product rule gives:
d[e^(2x)y]/dx = e^(3x)
Now integrate and solve for y:
e^(2x)y = ∫e^(3x) dx
e^(2x)y = e^(3x)/3 + C
y = e^x/3 + Ce^(-2x)
Check the solution
Differentiate y, then add 2y:
dy/dx = e^x/3 – 2Ce^(-2x)
dy/dx + 2y
= e^x/3 – 2Ce^(-2x) + 2e^x/3 + 2Ce^(-2x)
= e^x
The terms containing C cancel. That cancellation is a quick sign that the answer is on track. OpenStax explains first-order linear equations and the integrating-factor method in more detail.
Example 3: Solve a Homogeneous Differential Equation
Find the general solution to:
dy/dx = (x + y)/x
First rewrite the right side:
dy/dx = 1 + y/x
It depends on y/x, which suggests the substitution y = vx. Here, v is a function of x.
Substitute and integrate
Apply the product rule to y = vx:
dy/dx = v + x(dv/dx)
Replace y and dy/dx in the original equation:
v + x(dv/dx) = 1 + v
x(dv/dx) = 1
dv/dx = 1/x
Integrate, then replace v with y/x:
v = ln|x| + C
y/x = ln|x| + C
y = x(ln|x| + C)
This solution applies on an interval where x > 0 or x < 0. We cannot include x = 0 because the original equation divides by x.
Check the solution
Differentiate using the product rule:
y = x(ln|x| + C)
dy/dx = ln|x| + C + 1
Substitute the solution into the original right side:
(x + y)/x
= 1 + y/x
= 1 + ln|x| + C
Both sides agree. The substitution may look longer than the earlier methods, but recognizing the ratio y/x makes the steps straightforward.
What These Worked Solutions Have in Common
Each method changes the equation into something you can integrate. Separation does it by rearranging terms. An integrating factor creates a product derivative. The y = vx substitution turns a ratio into a new variable.
I also keep track of restrictions before rearranging anything. Dividing by y or x can exclude values from a calculation. A final answer must satisfy the original equation, not just the rewritten one.
If you are reviewing algebra alongside these steps, quadratic equations are useful practice for rearranging and checking solutions.
Frequently Asked Questions
1. What is the easiest way to recognize a separable equation?
Try rearranging it into the form g(y) dy = f(x) dx.
2. When should I use an integrating factor?
Use one for a first-order linear equation written as dy/dx + P(x)y = Q(x).
3. Why do homogeneous differential equations use y = vx?
When the right side depends on y/x, this substitution can reduce the equation to a separable form.
4. How can I check a differential equation solution?
Differentiate your proposed y and substitute both y and dy/dx into the original equation.
The Answer Isn’t Finished Until It Checks Out
When I practice solved examples of differential equations, I write the method beside the equation before doing any integration. That small step keeps me from applying a familiar technique to the wrong shape.
Try covering the solutions above and reworking one example of each type. Then differentiate your answers. A successful substitution check tells you more than reaching a plausible-looking final line.
Leave a Reply