First-Order Differential Equations: Types and Methods

What Are First-Order Differential Equations?

First-order differential equations are equations that involve an unknown function and its first derivative, but no second or higher derivatives. The general form is $F(x, y, y') = 0$, and most first courses work with the solved form

$$\frac{dy}{dx} = f(x, y),$$

where:

• $x$ is the independent variable (often time $t$ in applications),

• $y$ is the unknown function of $x$,

• $\dfrac{dy}{dx}$ is its first derivative, the rate of change,

• $f(x, y)$ is a given rule that says what the slope is at every point $(x, y)$.

The "first" refers to the order, the highest derivative present; the order and degree of a differential equation page explains how to read it. A first-order equation's general solution carries one arbitrary constant, so a single starting value is enough to fix one solution. These equations are the simplest branch of ordinary differential equations, and they are where every method you will use later first appears.

Geometrically, $\dfrac{dy}{dx} = f(x, y)$ assigns a slope to each point of the plane, and a solution is a curve that follows those slopes everywhere. Algebraically, a solution is a function that makes the equation true when you substitute it. Both views come back in the examples.

How Do You Tell If A First-Order Equation Is Linear Or Nonlinear?

A first-order equation is linear when it can be rearranged into the standard form

$$\frac{dy}{dx} + P(x),y = Q(x),$$

with $P$ and $Q$ functions of $x$ alone. That question, "How to determine if a first order differential equation is linear or non-linear", comes up on Quora because the test is about $y$, not about $x$. Run three checks on the dependent variable:

• Power one only. $y$ and $y'$ appear to the first power: no $y^2$, $\sqrt{y}$ or $(y')^2$.

• No products of $y$ with $y'$. A term such as $y,y'$ makes the equation nonlinear.

• No functions of $y$. $\sin y$, $e^{y}$ or $\ln y$ break linearity, even though $\sin x$, $e^{x}$ and $\ln x$ are fine.

The second row is the one that trips people. The functions of $x$ can be as complicated as you like; linearity only cares what happens to $y$.

What Are The Main Types Of First-Order Differential Equations?

Five types cover almost every first-order equation in a calculus or first ODE course: separable, linear, exact, homogeneous and Bernoulli. Each has a test you can run in a few seconds and a method that the test points to.

Two more names appear in longer courses. Autonomous equations $\dfrac{dy}{dx} = f(y)$ are a special case of separable ones, and Riccati and Clairaut equations are nonlinear types you will meet after these five.

One warning before choosing: the types overlap. The equation $x,\dfrac{dy}{dx} = 2y$ is separable, linear and homogeneous all at once, and every route gives $y = Cx^2$. When an equation fits several types, take the cheapest door, which is almost always separation.

How Do You Decide Which Method To Use?

Run the tests in order of effort and stop at the first one that passes. Separation is the fastest method, so it goes first; exactness takes partial derivatives, so it comes later.

1. Is it separable? Write $\dfrac{dy}{dx}$ alone on one side. If the right side factors as $g(x),h(y)$, separate.

2. Is it linear? If $y$ and $y'$ appear only to power one, put it in standard form and use an integrating factor.

3. Is it Bernoulli? If it would be linear except for a single $y^n$ term, substitute $v = y^{1-n}$.

4. Is it homogeneous? If replacing $x$ by $tx$ and $y$ by $ty$ leaves $f(x, y)$ unchanged, substitute $y = vx$.

5. Is it exact? Write it as $M,dx + N,dy = 0$ and compare $M_y$ with $N_x$.

6. Nearly exact? If $M_y \neq N_x$, look for an integrating factor that depends on $x$ alone or $y$ alone.

7. None of the above? Sketch the direction field or use a numerical method; many first-order equations have no formula solution at all.

The order matters less than the habit of testing at all. Most wrong answers in this topic come from reaching for a method before checking which type the equation is.

How Do You Solve A First-Order Linear Differential Equation?

You solve a first-order linear equation by putting it in standard form, multiplying through by the integrating factor $\mu(x) = e^{\int P(x),dx}$, and integrating. The factor is built so that the left side becomes the single derivative $\dfrac{d}{dx}\big[\mu y\big]$:

$$\frac{d}{dx}\big[\mu(x),y\big] = \mu(x),Q(x)$$

$$y = \frac{1}{\mu(x)}\left(\int \mu(x),Q(x),dx + C\right)$$

The integrating factor method page derives this from the product rule and works several examples, and the linear differential equations page covers the standard form in depth. For a hub, two facts matter most:

• Standard form first. If the coefficient of $y'$ is not $1$, divide by it before reading off $P(x)$.

• There is a second route. Variation of parameters solves $y' + Py = 0$ first, giving $y = Ce^{-\int P,dx}$, then lets $C$ become a function $u(x)$. It produces the same answer and generalises to higher-order equations.

Those two bullets, plus one integration, are the whole answer to the Brainly question "How do you solve this first-order linear differential equation?". The first bullet is where most attempts go wrong: for $x,y' + 2y = x^2$, the correct $P$ is $\dfrac{2}{x}$, not $2$.

How Do You Solve First-Order Nonlinear Differential Equations?

You solve a nonlinear first-order equation by finding a type it belongs to that turns it into something you can integrate: separable, exact, homogeneous or Bernoulli. That answers the Quora question "How to solve first order nonlinear differential equations": three of the five standard types are designed for nonlinear equations.

• Separable equations handle nonlinear terms in $y$ directly, because $h(y)$ moves to the other side; $\dfrac{dy}{dx} = x,y^2$ is nonlinear but separates at once.

• Homogeneous equations become separable after $y = vx$, however tangled $f\left(\dfrac{y}{x}\right)$ looks.

• Exact equations are often nonlinear and are solved as level curves $F(x, y) = C$.

• Bernoulli equations become linear after one substitution.

When none of these fits, the honest answer is that most nonlinear equations have no closed-form solution. You then read the behaviour from a slope field or compute values numerically, which is what engineering software does every day.

What Is A Bernoulli Equation And How Do You Solve It?

A Bernoulli equation is a first-order equation of the form

$$\frac{dy}{dx} + P(x),y = Q(x),y^n,$$

where $n$ is any real number. For $n = 0$ it is already linear, and for $n = 1$ it is linear and separable, so the interesting cases are every other $n$, where the $y^n$ term makes it nonlinear.

The substitution $v = y^{1-n}$ turns it into a linear equation. Divide through by $y^n$ and differentiate $v$ with the chain rule:

$$y^{-n},\frac{dy}{dx} + P(x),y^{1-n} = Q(x)$$

$$\frac{dv}{dx} = (1 - n),y^{-n},\frac{dy}{dx}$$

$$\frac{1}{1-n},\frac{dv}{dx} + P(x),v = Q(x)$$

$$\frac{dv}{dx} + (1 - n)P(x),v = (1 - n)Q(x)$$

The last line is linear in $v$, so the integrating factor finishes the job; then convert back with $y = v^{\frac{1}{1-n}}$. Dividing by $y^n$ assumes $y \neq 0$, so for $n > 0$ check $y = 0$ separately: it always solves the original equation.

The equation is named after Jacob Bernoulli, who discussed it in 1695. Gottfried Leibniz published the earliest solution that same year, Jacob Bernoulli published his own in 1696, and the substitution above is the standard modern route. The best-known Bernoulli equation is the logistic population model $\dfrac{dP}{dt} = rP\left(1 - \dfrac{P}{K}\right)$, which is Bernoulli with $n = 2$ (and also separable).

How Do You Solve A Homogeneous First-Order Equation?

You solve a homogeneous equation $\dfrac{dy}{dx} = f\left(\dfrac{y}{x}\right)$ by substituting $y = vx$, so $\dfrac{dy}{dx} = v + x,\dfrac{dv}{dx}$, which makes the equation separable in $v$ and $x$. The Quora question "What are examples of first order homogeneous differential equations? How are they solved?" has a short answer: any equation whose right side depends only on the ratio $\dfrac{y}{x}$, such as

$$\frac{dy}{dx} = \frac{x + y}{x}, \qquad \frac{dy}{dx} = \frac{x^2 + y^2}{xy}, \qquad \frac{dy}{dx} = \frac{y}{x} + \tan\frac{y}{x}.$$

The test is $f(tx, ty) = f(x, y)$ for every $t \neq 0$. For $\dfrac{x^2 + y^2}{xy}$, replacing $x$ and $y$ by $tx$ and $ty$ puts $t^2$ on top and bottom, and they cancel. Example 4 below works a full one, taken from a Math Stack Exchange thread, and the homogeneous differential equations page has more.

A naming warning: "homogeneous" also describes linear equations with zero right side, $y' + P(x)y = 0$. The two meanings are unrelated, and textbooks use both.

What Does A First-Order Equation Look Like Geometrically?

A first-order equation $\dfrac{dy}{dx} = f(x, y)$ is a slope field: at every point it draws a short segment with slope $f(x, y)$, and each solution is a curve that runs along those segments. The general solution is the whole family of such curves, one for each value of the constant.

• Initial values pick one curve. Choosing $y(x_0) = y_0$ selects the curve through that point; pairing an equation with that value makes an initial value problem.

• Equilibrium solutions are flat lines. Where $f(x, y) = 0$ for all $x$ at a fixed value $y = c$, the horizontal line $y = c$ is a solution. For $\dfrac{dA}{dt} = 2 - \dfrac{A}{25}$ that line is $A = 50$.

• Nearby curves show stability. If neighbouring curves bend toward an equilibrium, it is stable; if they peel away, it is unstable.

Examples Of First-Order Differential Equations

Each example names its type first, because recognising the type is the skill this page is about. Every answer is checked by substitution.

Example 1

Separable. Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$.

$$\frac{dy}{y} = 2x,dx$$

$$\ln|y| = x^2 + C$$

$$y = Ae^{x^2}$$

$$y(0) = A = 3$$

Check: $y = 3e^{x^2}$ gives $\dfrac{dy}{dx} = 6xe^{x^2} = 2x \cdot 3e^{x^2}$.

Final answer: $y = 3e^{x^2}$.

Example 2

Solve $\dfrac{dy}{dx} = x + y$ with $y(0) = 1$.

The right side has both letters, so the reflex is to separate:

$$\frac{dy}{y} = x,dx$$

$$\ln|y| = \frac{x^2}{2} + C$$

$$y = Ae^{x^2/2}$$

Test it: $\dfrac{dy}{dx} = Axe^{x^2/2}$, but $x + y = x + Ae^{x^2/2}$. These are not equal, so the "solution" fails. The first line was the error: dividing by $y$ only separates a product $x \cdot y$, and the right side here is a sum $x + y$.

Run the linearity test instead. $y$ appears to power one, so the equation is linear:

$$\frac{dy}{dx} - y = x$$

$$\mu = e^{\int (-1),dx} = e^{-x}$$

$$\frac{d}{dx}\big[e^{-x}y\big] = xe^{-x}$$

$$e^{-x}y = -(x + 1)e^{-x} + C$$

$$y = Ce^{x} - x - 1$$

$$y(0) = C - 1 = 1, \quad C = 2$$

Check: $y = 2e^{x} - x - 1$ gives $\dfrac{dy}{dx} = 2e^{x} - 1$ and $x + y = 2e^{x} - 1$.

Final answer: $y = 2e^{x} - x - 1$.

Example 3

Exact. Solve $(2xy + \cos x),dx + (x^2 + 3y^2),dy = 0$.

$$M = 2xy + \cos x, \qquad N = x^2 + 3y^2$$

$$\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}$$

The equation is exact, so look for $F(x, y)$ with $F_x = M$:

$$F = \int (2xy + \cos x),dx = x^2 y + \sin x + h(y)$$

$$F_y = x^2 + h'(y) = x^2 + 3y^2, \quad h(y) = y^3$$

Check: $F = x^2 y + \sin x + y^3$ has $F_x = 2xy + \cos x$ and $F_y = x^2 + 3y^2$.

Final answer: $x^2 y + \sin x + y^3 = C$.

Example 4

Homogeneous. Solve $(x^2 - 2y^2),dx + xy,dy = 0$.

This one comes from the Math Stack Exchange thread "Help in solving ODE $(x^2-2y^2)dx+xydy=0$ which general solution includes modulus". Rewrite and substitute $y = vx$:

$$\frac{dy}{dx} = \frac{2y^2 - x^2}{xy} = \frac{2v^2 - 1}{v}$$

$$v + x,\frac{dv}{dx} = \frac{2v^2 - 1}{v}$$

$$x,\frac{dv}{dx} = \frac{v^2 - 1}{v}$$

$$\frac{v,dv}{v^2 - 1} = \frac{dx}{x}$$

$$\frac{1}{2}\ln|v^2 - 1| = \ln|x| + c$$

$$v^2 - 1 = Kx^2$$

The absolute values, which caused the thread's confusion, disappear into the constant: $K$ can be any real number, positive or negative. Replace $v$ by $\dfrac{y}{x}$:

$$y^2 = x^2 + Kx^4$$

Check: SymPy confirms $y = \pm\sqrt{x^2 + Kx^4}$ satisfies the equation for every $K$. The value $K = 0$ gives $y = \pm x$, the two lines lost when dividing by $v^2 - 1$.

Final answer: $y^2 = x^2 + Kx^4$.

Example 5

Bernoulli. Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 y^2$ for $x > 0$.

Here $n = 2$, so $v = y^{-1}$ and the linear equation is $v' + (1 - 2)\dfrac{1}{x}v = (1 - 2)x^2$:

$$\frac{dv}{dx} - \frac{v}{x} = -x^2$$

$$\mu = e^{-\ln x} = \frac{1}{x}$$

$$\frac{d}{dx}\left[\frac{v}{x}\right] = -x$$

$$\frac{v}{x} = -\frac{x^2}{2} + \frac{C}{2}$$

$$v = \frac{Cx - x^3}{2}$$

$$y = \frac{2}{Cx - x^3}$$

The constant was written as $\dfrac{C}{2}$ only to keep the final fraction clean. Check: SymPy's dsolve returns $y = \dfrac{2}{x(C - x^2)}$, the same family, and substitution leaves a residual of $0$. The solution $y = 0$ also works and is not in the family.

Final answer: $y = \dfrac{2}{Cx - x^3}$, plus $y = 0$.

Example 6

A 100-litre tank holds 20 kg of dissolved salt. Brine with 0.5 kg of salt per litre flows in at 4 litres per minute, and the mixed solution drains at 4 litres per minute. Find the salt $A(t)$ after $t$ minutes, the amount after 25 minutes, and when it reaches 45 kg.

Salt enters at $0.5 \times 4 = 2$ kg per minute and leaves at $\dfrac{A}{100} \times 4 = \dfrac{A}{25}$ kg per minute:

$$\frac{dA}{dt} = 2 - \frac{A}{25}, \quad A(0) = 20$$

The equation is both separable and linear. Treating it as linear, $\mu = e^{t/25}$:

$$\frac{d}{dt}\big[e^{t/25}A\big] = 2e^{t/25}$$

$$e^{t/25}A = 50e^{t/25} + C$$

$$A = 50 + Ce^{-t/25}$$

$$A(0) = 50 + C = 20, \quad C = -30$$

$$A(25) = 50 - 30e^{-1} \approx 38.9636$$

$$50 - 30e^{-t/25} = 45, \quad t = 25\ln 6 \approx 44.7940$$

Check: $\dfrac{dA}{dt} = \dfrac{6}{5}e^{-t/25}$ and $2 - \dfrac{A}{25} = 2 - 2 + \dfrac{6}{5}e^{-t/25}$.

Final answer: $A(t) = 50 - 30e^{-t/25}$; about 38.96 kg after 25 minutes; 45 kg after about 44.79 minutes.

Why Do First-Order Differential Equations Matter?

First-order equations matter because so many processes follow one rule: the rate of change depends on the current amount. Once you can write that rule down, the equation predicts the future from a single measurement of the present.

• Growth and decay. $\dfrac{dy}{dt} = ky$ governs radioactive decay, continuously compounded interest and unchecked population growth, the core of exponential growth and decay.

• Cooling and heating. Newton's law of cooling is first-order linear, as is the salt balance in Example 6.

• Motion with drag. A falling object with air resistance proportional to speed obeys a first-order linear equation for its velocity, worked in OpenStax Calculus Volume 2.

• Circuits. The current in a resistor and inductor circuit satisfies $L,\dfrac{di}{dt} + Ri = V$.

• Limited growth. The logistic equation, a Bernoulli equation, caps growth at a carrying capacity.

Knowing the type pays off here. The same mixing equation, read as separable or as linear, gives the same answer, so an engineer can pick whichever method is faster.

Where Do Solutions Go Off The Rails?

Most mistakes with first-order equations happen before any integration, at the moment of deciding what kind of equation it is.

Mistake 1: Separating A Sum

Where it slips in: Equations such as $y' = x + y$ or $y' = x^2 + y^2$, where both variables appear on the right.

Don't do this: Divide by $y$ and multiply by $dx$ as though the right side were a product.

The correct way: Separate only when the right side factors as $g(x),h(y)$. Otherwise test for linear, Bernoulli or homogeneous.

The rusher makes this one: they see $x$ and $y$ and start separating before reading the operation between them. Asking "is that a plus or a times?" before the first line of working stops it.

Mistake 2: Reading P Before Standard Form

Where it slips in: Linear equations that arrive as $x,y' + 2y = x^2$ or $3y' - 6y = e^x$.

Don't do this: Take $P = 2$ or $P = -6$ straight from the equation.

The correct way: Divide by the coefficient of $y'$ first: $P = \dfrac{2}{x}$ in the first case and $P = -2$ in the second.

Mistake 3: Calling An Equation Linear Because It Looks Simple

Where it slips in: Short equations such as $y,y' = x$ or $y' = \cos y$.

Don't do this: Reach for the integrating factor because the equation is first-order and short.

The correct way: Run the three linearity checks on $y$. Both examples fail, and both turn out to be separable.

Mistake 4: Losing Solutions When Dividing

Where it slips in: Dividing by $y$ in separable equations, by $y^n$ in Bernoulli equations, or by $v^2 - 1$ in homogeneous ones.

Don't do this: Divide and move on without asking whether the divisor can be zero.

The correct way: Test each excluded value: $y = 0$ in Examples 1 and 5, and $y = \pm x$ in Example 4, are genuine solutions.

The student who memorised one method per chapter is the one these mistakes catch hardest, because a test asks which method, not which chapter. The fix is the seven-step checklist above, run every time, even when the answer feels obvious.

Practice Problems On First-Order Differential Equations

1. Solve $\dfrac{dy}{dx} = xy^2$ with $y(0) = 1$.

2. Find the general solution of $\dfrac{dy}{dx} + 2y = 4x$.

3. Show that $(2xy + 3),dx + (x^2 + 4y),dy = 0$ is exact and solve it.

4. Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$ for $x > 0$.

5. Solve the Bernoulli equation $\dfrac{dy}{dx} - y = -y^2$ with $y(0) = \dfrac{1}{2}$.

6. Name the type of each equation: (a) $y' = \sin x,\cos y$; (b) $y' + x^2 y = e^{x}$; (c) $y' = \dfrac{x^2 + y^2}{xy}$; (d) $y' + y = y^4$.

Answer to Question 1: $y = \dfrac{2}{2 - x^2}$ (separable).

Answer to Question 2: $y = 2x - 1 + Ce^{-2x}$ (linear, $\mu = e^{2x}$).

Answer to Question 3: $M_y = 2x = N_x$; the solution is $x^2 y + 3x + 2y^2 = C$.

Answer to Question 4: $y = x\ln x + Cx$ (homogeneous, or linear as $y' - \dfrac{y}{x} = 1$).

Answer to Question 5: $y = \dfrac{1}{1 + e^{-x}}$ (Bernoulli with $n = 2$, $v = y^{-1}$).

Answer to Question 6: (a) separable; (b) linear; (c) homogeneous; (d) Bernoulli with $n = 4$.

The Short Version

• First-order differential equations involve only $y'$ and need one starting value to fix one solution.

• Five types cover most problems: separable, linear, exact, homogeneous and Bernoulli, each with a quick test.

• Test in order of effort, separation first, and stop at the first test that passes.

• Linearity is about $y$ only: power one, no $y$ inside functions, no $y,y'$ products.

• Check every division for lost solutions, and verify every answer by substitution.

Where To Go From Here

Work through problems 1 to 6 in order; if problem 6 feels like guessing, return to the decision checklist, and if problem 5 stalls, reread the Bernoulli derivation.

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