Linear differential equations (first order)

A first-order differential equation is called linear when the dependent variable and its derivative appear only to the first power, with no products like $y\,\dfrac{dy}{dx}$ or terms like $y^2$ . Such an equation usually cannot be split into " $x$ -stuff on one side, $y$ -stuff on the other" the way a separable equation can. Instead, CBSE teaches a clever trick: multiply the whole equation by a special function called the integrating factor (IF), chosen precisely so that the left-hand side collapses into the derivative of a single product. Once that happens, the equation solves itself by direct integration. This is the third and last solving technique in the Class 12 Differential Equations chapter, after variables separable and homogeneous equations. The syllabus restricts you to two standard forms — linear in $y$ , or linear in $x$ — and one job: identify the form, build the IF, plug into a fixed formula, and integrate. Exact equations and Bernoulli's equation are outside this syllabus.

The core insight

Multiplying $\dfrac{dy}{dx}+py=q$ by the integrating factor $e^{\int p\,dx}$ turns the left side into $\dfrac{d}{dx}(y\cdot\mathrm{IF})$ , so the whole equation integrates directly to give $y$ .

The core idea

Recognise the standard form. A first-order linear equation in $y$ looks like

\frac{dy}{dx} + py = q,

where $p$ and $q$ are constants or functions of $x$ only — never of $y$ . If the equation you are given is not already written this way (coefficient of $\dfrac{dy}{dx}$ not equal to $1$ ), divide through first; reading off $p$ and $q$ from the un-simplified equation is the single most common source of error.

Build the integrating factor. The IF is defined as

\mathrm{IF} = e^{\int p\,dx}.

where: $p$ is the coefficient function from the standard form and $\int p\,dx$ is computed without a constant of integration (any constant would cancel out later, so it is dropped by convention).

Read off the solution formula. Multiplying the standard equation by $\mathrm{IF}$ and integrating both sides gives the general solution directly:

y \cdot \mathrm{IF} = \int (q \cdot \mathrm{IF})\,dx + C.

where: $C$ is the arbitrary constant of integration, fixed by an initial condition if one is given.

Mind the mirror form. Some equations are more naturally linear in $x$ (treating $x$ as the dependent variable), with standard form $\dfrac{dx}{dy}+px=q$ , where $p,q$ are functions of $y$ only. The same machinery applies with $x$ and $y$ swapped: $\mathrm{IF}=e^{\int p\,dy}$ and $x\cdot\mathrm{IF} = \int (q\cdot\mathrm{IF})\,dy + C$ . Deciding which variable the equation is linear in — before touching the formula — is the real skill being tested. A quick giveaway: if isolating $\dfrac{dy}{dx}$ leaves a coefficient or a right-hand side that still depends on $y$ , the equation is not linear in $y$ at all, and you should instead try writing it as $\dfrac{dx}{dy}+px=q$ and check whether $p,q$ come out depending only on $y$ .

Why this is different from the earlier two methods. Variables separable equations let you sort every $x$ -term to one side and every $y$ -term to the other before integrating; homogeneous equations use the substitution $y=vx$ to reduce to a separable one. A linear equation of the form $\dfrac{dy}{dx}+py=q$ generally cannot be separated or reduced by $y=vx$ at all — the $y$ and $x$ dependence are tangled together through the sum $py$ . The integrating factor is not a substitution that changes variables; it is a multiplier that reshapes the equation itself so that ordinary integration becomes possible, which is why it is introduced as a genuinely new technique rather than a variant of the first two.

Figure

A flow diagram showing the four steps from a linear differential equation to its general solution

Derivation AWhy the integrating factor makes the equation directly integrable

The goal is to show, from the product rule alone, that multiplying $\dfrac{dy}{dx}+py=q$ by $\mathrm{IF}=e^{\int p\,dx}$ turns the left-hand side into the exact derivative $\dfrac{d}{dx}(y\cdot\mathrm{IF})$ , which is the entire reason the method works and the fact every student is expected to justify, not just apply.

Step 1. Multiply the standard equation $\dfrac{dy}{dx}+py=q$ throughout by $\mathrm{IF}=e^{\int p\,dx}$ . This gives $\mathrm{IF}\cdot\dfrac{dy}{dx} + p\cdot\mathrm{IF}\cdot y = q\cdot\mathrm{IF}$ . Nothing has been solved yet — this is just a relabelling of the same equation with every term scaled by the same function of $x$ .

Step 2. Differentiate $\mathrm{IF}$ itself using the chain rule. Since $\mathrm{IF}=e^{\int p\,dx}$ , $\dfrac{d(\mathrm{IF})}{dx} = e^{\int p\,dx}\cdot\dfrac{d}{dx}\left(\int p\,dx\right) = \mathrm{IF}\cdot p$ , because the derivative of $\int p\,dx$ with respect to $x$ is simply $p$ by the fundamental theorem of calculus.

Step 3. Now apply the product rule to $y\cdot\mathrm{IF}$ : $\dfrac{d}{dx}\big(y\cdot\mathrm{IF}\big) = \dfrac{dy}{dx}\cdot\mathrm{IF} + y\cdot\dfrac{d(\mathrm{IF})}{dx} = \mathrm{IF}\cdot\dfrac{dy}{dx} + y\cdot(\mathrm{IF}\cdot p)$ . Compare this term by term with the left-hand side obtained in Step 1 — they are identical, so $\mathrm{IF}\cdot\dfrac{dy}{dx} + p\cdot\mathrm{IF}\cdot y = \dfrac{d}{dx}\big(y\cdot\mathrm{IF}\big)$ exactly.

Step 4. Substitute this back into the multiplied equation from Step 1: the left side becomes $\dfrac{d}{dx}(y\cdot\mathrm{IF})$ , so the whole equation reads $\dfrac{d}{dx}\big(y\cdot\mathrm{IF}\big) = q\cdot\mathrm{IF}$ . Integrating both sides with respect to $x$ removes the derivative on the left and leaves $y\cdot\mathrm{IF} = \int(q\cdot\mathrm{IF})\,dx + C$ , the general solution formula, now shown to follow from nothing more than the product rule and the definition of $\mathrm{IF}$ .

Worked example

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

Step 1. Identify the standard form. The equation already matches $\dfrac{dy}{dx}+py=q$ with $p=\dfrac{1}{x}$ and $q=x^2$ ; no rearrangement is needed since the coefficient of $\dfrac{dy}{dx}$ is already $1$ .

Step 2. Compute the integrating factor: $\mathrm{IF} = e^{\int p\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x$ (since $x>0$ , no absolute value is needed).

Step 3. Apply the solution formula: $y\cdot x = \int\left(x^2\cdot x\right)dx + C = \int x^3\,dx + C = \dfrac{x^4}{4}+C$ .

Step 4. Solve for $y$ : $xy = \dfrac{x^4}{4}+C \implies \boxed{y = \dfrac{x^3}{4} + \dfrac{C}{x}}$ . Checking by differentiating confirms $\dfrac{dy}{dx}+\dfrac{y}{x} = \dfrac{3x^2}{4}-\dfrac{C}{x^2}+\dfrac{x^2}{4}+\dfrac{C}{x^2} = x^2$ , which matches the original equation.

Competitive stretch

If the same equation instead carried an initial condition $y=1$ when $x=2$ , you would substitute directly into the general solution rather than re-deriving anything: $1 = \dfrac{2^3}{4}+\dfrac{C}{2} = 2 + \dfrac{C}{2} \implies C = -2$ , giving the particular solution $y=\dfrac{x^3}{4}-\dfrac{2}{x}$ . This is the standard "general solution first, then plug in the condition" pattern examiners expect — never try to build the condition into the integrating factor itself.

Common traps

Reading off $p$ and $q$ before reducing to standard form. If the coefficient of $\dfrac{dy}{dx}$ is not $1$ , dividing through is compulsory first; skipping this step gives a wrong integrating factor.
Choosing the wrong variable to be linear in. An equation linear in $x$ (form $\dfrac{dx}{dy}+px=q$ ) forced into the $y$ -form produces a non-linear mess this method cannot solve.
Mishandling $\int\frac{1}{x}\,dx$ and the exponent-log simplification. This integral is $\ln|x|$ , so $\mathrm{IF}=e^{\ln|x|}=|x|$ in general; the absolute value is dropped only once a domain restriction like $x>0$ is stated.
Applying the initial condition to the original equation instead of the integrated one. $C$ must come from $y\cdot\mathrm{IF}=\ldots$ , never from $\dfrac{dy}{dx}+py=q$ directly.

Ask the tutor: if computing $\int p\,dx$ or simplifying $e^{\int p\,dx}$ feels like the tricky part, ask your AI tutor to walk through it step by step for your specific equation.

Why does the integrating factor have no plus C in its own exponent?

Because any constant $k$ added inside $\int p\,dx$ would appear as a multiplying factor $e^k$ on both sides of $y\cdot\mathrm{IF}=\int(q\cdot\mathrm{IF})dx+C$ , and dividing it out leaves the same relationship between $y$ and $x$ . It changes nothing about the final solution, so by convention the constant is simply dropped when computing $\int p\,dx$ for the integrating factor, keeping the algebra as clean as possible.

How do I tell whether an equation is linear in y or linear in x?

Look at where the derivative sits and what it is being divided or multiplied by. If you can isolate $\dfrac{dy}{dx}$ and the rest of the equation involves only $x$ (and $y$ to the first power, unmultiplied by any $x$ -function times $y$ 's own derivative in a nonlinear way), it is linear in $y$ . If instead $\dfrac{dx}{dy}$ isolates cleanly with coefficients depending only on $y$ , treat $x$ as the dependent variable instead. Try the $y$ -form first since it is more common in the syllabus; if $p$ or $q$ end up depending on $y$ , switch to the $x$ -form.

What happens if I forget the constant of integration C?

You lose the general solution and get only one member of an entire family of solutions — usually not the one the question wants. Since first-order equations always produce one arbitrary constant, $C$ must appear in $y\cdot\mathrm{IF}=\int(q\cdot\mathrm{IF})dx+C$ and stay until either the question asks only for the general solution (leave it as $C$ ) or gives an initial condition (solve for its specific numeric value).

Linear differential equations (first order)

The core idea

Worked example

Competitive stretch

Common traps

Before this