How To Solve First Order Linear Differential Equation

The Equation That Shows Up Everywhere (And How to Actually Solve It)

You encounter a first order linear differential equation and your stomach drops. It looks intimidating — a derivative, a function, and some coefficient thrown in for good measure. Plus, the real challenge isn't the math itself. But here's the thing: this is one of the most structured, repeatable problem types in all of calculus. Once you internalize the pattern, it becomes almost mechanical. It's knowing which step comes first and why.

This guide walks through the entire process — from recognizing the equation to applying the method correctly, avoiding the traps that trip up most students, and building intuition along the way.

What Is a First Order Linear Differential Equation

A first order differential equation involves a derivative of the first degree — meaning you're dealing with dy/dx (or dy/dt, or whatever variable you have), but no second derivatives or higher. In practice, "Linear" means the dependent variable and its derivative appear only to the first power. They aren't squared, they aren't inside a sine function, and they aren't multiplied together.

The general form looks like this:

dy/dx + P(x)y = Q(x)

That's it. P(x) and Q(x) are functions of x — they can be constants, polynomials, exponentials, or anything else that depends on x. The key signature is that y and dy/dx sit alone, each raised to the first power, and nothing in the equation involves y × dy/dx or y².

What Makes It "First Order" and "Linear"

"First order" just means the highest derivative present is the first derivative. "Linear" means the equation is linear in y and its derivatives. If you see something like dy/dx + y² = x, that's first order but not linear — because of the y² term. Similarly, dy/dx + y·sin(y) = 0 is nonlinear. The linearity condition is what makes the integrating factor method work so cleanly.

Why This Form Matters

The standard form dy/dx + P(x)y = Q(x) isn't just a convention. It's the shape that tells you exactly what to do next. If your equation doesn't look like this yet, your first step is to rearrange it. That single act of putting it into standard form solves half the problem before you even start calculating.

Why It Matters (Beyond the Exam)

First order linear differential equations model a surprising number of real-world situations. Consider this: mixing problems where a chemical concentration changes over time in a tank with varying inflow? So naturally, that can land in this form. Population growth with a carrying capacity and a time-dependent harvesting rate? Often linear and first order. Even certain electrical circuits — like an RC circuit with a time-varying voltage source — reduce to a first order linear equation.

Understanding how to solve these equations gives you a tool that applies across physics, engineering, economics, and biology. The technique itself — the integrating factor — also generalizes to higher-order linear equations and systems, so getting comfortable with it now pays dividends later.

How to Solve a First Order Linear Differential Equation

The method has a clear sequence of steps. Here's the full breakdown.

Step 1: Put the Equation into Standard Form

Before you do anything else, make sure your equation matches dy/dx + P(x)y = Q(x). If it doesn't, algebraically manipulate it until it does.

Here's one way to look at it: suppose you're given 2(dy/dx) + 3xy = eˣ. Which means divide everything by 2 to get dy/dx + (3x/2)y = eˣ/2. Now P(x) = 3x/2 and Q(x) = eˣ/2. That's standard form.

This step is non-negotiable. If you skip it or do it carelessly, every calculation that follows will be wrong.

Step 2: Find the Integrating Factor

The integrating factor, usually written μ(x), is the engine of this entire method. You compute it using the formula:

μ(x) = e^∫P(x) dx

That's the exponential of the integral of P(x). No constant of integration is needed here — any constant you'd add would just cancel out later, so you can safely ignore it.

Going back to our example, P(x) = 3x/2, so ∫P(x) dx = (3x²)/4. The integrating factor is μ(x) = e^(3x²/4).

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Step 3: Multiply the Entire Equation by the Integrating Factor

Take the standard form equation and multiply both sides by μ(x). The left-hand side will magically collapse into the derivative of a product — specifically, d/dx[μ(x) · y].

This is not a coincidence. Think about it: it exploits the product rule in reverse. The integrating factor is constructed precisely so that this works. When you multiply dy/dx + P(x)y by μ(x), you get μ(x)dy/dx + μ(x)P(x)y, which equals d/dx[μ(x)y] by design.

In our example, after multiplying by e^(3x²/4), the left side becomes d/dx[e^(3x²/4) · y]. The right side becomes e^(3x²/4) · eˣ/2, which simplifies to (1/2)e^(3x²/4 + x).

Step 4: Integrate Both Sides

Now you have an equation of the form d/dx[μ(x)y] = μ(x)Q(x). Integrate both sides with respect to x.

The left side integrates to μ(x)y (since the integral of a derivative just gives back the original function, plus a constant). The right side requires you to compute ∫μ(x)Q(x) dx.

This integral is where the difficulty often lives. On top of that, in some cases it's clean. In others, it doesn't have a closed-form expression, and you leave it in integral notation — which is perfectly acceptable.

For our example, you'd write:

e^(3x²/4) · y = ∫(1/2)e^(3x²/4 + x) dx + C

If the integral on the right doesn't simplify nicely, you can express the solution as y = e^(-3x²/4) times that integral, plus the constant term. That's a valid, complete answer.

Step 5: Solve for y

Finally, divide both sides by μ(x) to isolate y. You now have your general solution.

y = (1/μ(x)) · [∫μ(x)Q(x) dx + C]

That's the formula in its fully general form. Every first order linear differential equation solution follows this exact structure.

A Second Example for Clarity

Consider

Consider the differential equation

[ \frac{dy}{dx} - \frac{2}{x}y = x^{3}. ]

Put it in standard form.
The equation already reads (\displaystyle \frac{dy}{dx}+P(x)y = Q(x)) with

[ P(x)=-\frac{2}{x},\qquad Q(x)=x^{3}. ]

Compute the integrating factor.
[ \mu(x)=\exp!\left(\int P(x),dx\right)=\exp!\left(\int -\frac{2}{x},dx\right) =\exp!\bigl(-2\ln|x|\bigr)=|x|^{-2}. ]
For (x>0) we can drop the absolute value and write (\mu(x)=x^{-2}).

Multiply the whole equation by (\mu(x)).
[ x^{-2}\frac{dy}{dx}-2x^{-3}y = x. ]
The left‑hand side is exactly (\displaystyle \frac{d}{dx}!\bigl(x^{-2}y\bigr)) by construction.

Integrate both sides.
[ \frac{d}{dx}!\bigl(x^{-2}y\bigr)=x \quad\Longrightarrow\quad x^{-2}y = \int x,dx = \frac{x^{2}}{2}+C, ]
where (C) is an arbitrary constant.

Solve for (y).
Multiply by (x^{2}):

[ y = x^{2}!\left(\frac{x^{2}}{2}+C\right)=\frac{x^{4}}{2}+Cx^{2}. ]

Thus the general solution of the original equation is

[ \boxed{,y=\dfrac{x^{4}}{2}+Cx^{2},}. ]

Conclusion

The integrating‑factor method transforms any first‑order linear differential equation into an exact derivative, allowing a straightforward integration. The five‑step recipe—standard form, integrating factor, multiplication, integration, and back‑substitution—works universally, whether the resulting integral is elementary or must be left in integral notation. Mastering this technique not only yields solutions to a broad class of problems in physics, engineering, and applied mathematics but also deepens intuition about how linear operators can be “undone” by clever multiplication. When faced with a new linear ODE, remember: put it in the right shape, build the integrating factor, let the product rule do the heavy lifting, integrate, and finally isolate (y). This systematic approach guarantees correctness and efficiency.