Differential Equations

Every calculus problem you have solved so far asked you to find a number, or a single function, that satisfies some equation. This chapter asks for something new: find a function $y = f(x)$ whose own derivative, or a combination of its derivatives, is prescribed to you in advance. The simplest possible instance is one you already know how to finish — $\dfrac{dy}{dx} = e^x$ is solved the moment you integrate both sides. Dress that idea up, generalise it, and give it a name, and you have the whole chapter in miniature. An equation that involves an unknown function together with its derivatives is called a differential equation, and this chapter builds the vocabulary to classify such equations and the machinery to solve the first order, first degree kind. Two numbers classify a differential equation before you attempt to solve it: its order (how many times the dependent variable was differentiated) and its degree (the power of the highest derivative, but only when the equation is a polynomial in its derivatives at all — otherwise degree is simply not defined). Solving a differential equation does not return one number the way solving $x^2 - 3x + 3 = 0$ does; it returns an entire family of functions, the general solution, carrying as many arbitrary constants as the order of the equation. Pin down those constants with an initial condition — a point the curve must pass through — and the family collapses to one particular solution. The chapter's real payload is three complementary methods for the first order, first degree case: variables separable, where the two variables can be herded onto opposite sides of the equals sign and integrated directly; homogeneous equations, tamed by the substitution $y = vx$ ; and linear equations, solved by multiplying through by an integrating factor that turns the left side into the derivative of a single product. Every exercise in this chapter is one of these three disguises, and learning to recognise which disguise you are looking at — before you touch a pencil to the algebra — is the actual skill being tested.

Key takeaways

A differential equation relates a dependent variable to the derivative or derivatives of itself with respect to one or more independent variables; restricting to a single independent variable gives an ordinary differential equation, the only kind this chapter studies.
The order of a differential equation is the order of its highest derivative; the degree is the highest power of that highest derivative, defined only when the equation is a polynomial equation in its derivatives.
A function that satisfies a differential equation is its solution; a solution carrying arbitrary constants equal in number to the order is the general solution (primitive), and a solution obtained by assigning particular values to those constants is a particular solution.
Solving $\dfrac{dy}{dx} = g(y)\,h(x)$ by variables separable: rewrite as $\dfrac{dy}{h(y)} = g(x)\,dx$ and integrate both sides independently.
A differential equation $\dfrac{dy}{dx} = F(x,y)$ is homogeneous exactly when $F$ is a homogeneous function of degree zero, i.e. $F(x,y) = g\!\left(\dfrac{y}{x}\right)$ ; the substitution $y = vx$ then reduces it to a variables separable equation in $v$ and $x$ .
A linear differential equation has the standard form $\dfrac{dy}{dx} + Py = Q$ with $P, Q$ functions of $x$ alone (or the mirror form $\dfrac{dx}{dy} + P_1 x = Q_1$ ); multiplying by the integrating factor $\text{I.F.} = e^{\int P\,dx}$ makes the left side an exact derivative.
The general solution of a linear equation is $y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\, dx + C$ , obtained by integrating both sides once the integrating factor has collapsed the left side to $\dfrac{d}{dx}\!\left(y \cdot \text{I.F.}\right)$ .
Verifying that a given function solves a differential equation is pure differentiation: differentiate the candidate the required number of times and substitute into the equation to confirm both sides match identically, for every value of the arbitrary constants.
Key idea: classify a first order first degree equation as separable, homogeneous, or linear by its algebraic form before solving, since each disguise has its own fixed recipe — separate the variables, substitute y = vx, or multiply by e to the integral of P dx.

The core insight

A differential equation relates a function to its own derivatives and is classified by order and degree before being solved; first order first degree equations are recognised as one of three standard forms — variables separable, homogeneous ( $y = vx$ ), or linear (integrating factor $e^{\int P\,dx}$ ) — each of which reduces the equation to something directly integrable.

What this chapter covers

The chapter builds in a fixed order, and it pays to keep the stages distinct in your head. It opens with classification — Sections 9.2.1 and 9.2.2 in the NCERT text — where every differential equation is assigned an order (from its highest derivative) and, when the equation is polynomial in its derivatives, a degree (the power that derivative is raised to). Next comes the idea of a solution itself (Section 9.3): unlike an algebraic equation, whose solution is a number, a differential equation's solution is a function, and the chapter draws a sharp line between the general solution (carrying arbitrary constants equal in number to the order) and a particular solution (those constants pinned to specific values by a given condition). The bulk of the chapter, Section 9.4, is a toolkit of three methods for first order, first degree equations, presented in the order you should try them when facing an unfamiliar problem. Variables separable (9.4.1) applies when the two variables can be physically separated onto opposite sides of the equation. Homogeneous equations (9.4.2) apply when the right-hand side is a ratio of same-degree expressions in $x$ and $y$ , tamed by writing $y = vx$ . Linear equations (9.4.3) apply when the dependent variable and its derivative appear only to the first power, unmultiplied by each other, solved with an integrating factor. The worked examples and derivations below trace this exact order — order and degree first, then general versus particular solutions, then the three methods each in turn — and the miscellaneous examples at the end of the chapter mix all three ideas together, which is exactly how competitive exams test the chapter.

The core ideas

Order and degree of a differential equation

An equation involving derivatives of a dependent variable with respect to an independent variable is a differential equation; when only one independent variable is involved, it is called an ordinary differential equation, and that is the only kind this chapter studies (equations with derivatives with respect to more than one independent variable are partial differential equations, outside this chapter's scope). NCERT's own shorthand for derivatives is worth adopting immediately: $y' = \dfrac{dy}{dx}$ , $y'' = \dfrac{d^2y}{dx^2}$ , $y''' = \dfrac{d^3y}{dx^3}$ , and for still higher derivatives, $y_n = \dfrac{d^n y}{dx^n}$ .

\text{Order} = \text{the order of the highest derivative of } y \text{ present in the equation.} \qquad (1)

where: order is read straight off the equation — count how many times $y$ has been differentiated in the highest term present, nothing more. $\dfrac{dy}{dx} = e^x$ has order one; $\dfrac{d^2y}{dx^2} + y = 0$ has order two. Degree is subtler, because it is only defined when the differential equation is a polynomial equation in its derivatives — meaning $y'$ , $y''$ , and so on appear only raised to non-negative integer powers, never inside a sine, a logarithm, a square root, or a denominator.

\text{Degree} = \text{the highest power (positive integer) of the highest order derivative,} \qquad (2)

where: the definition applies only when the equation is polynomial in its derivatives. An equation such as $\dfrac{dy}{dx} + \sin\!\left(\dfrac{dy}{dx}\right) = 0$ is not a polynomial equation in $y'$ at all — you cannot write $\sin(y')$ as a finite sum of integer powers of $y'$ — so its degree is simply not defined, not zero, not undefined-but-approximately-something. Order and degree, whenever degree exists, are always positive integers.

Equation	Order	Degree
$\dfrac{dy}{dx} = e^x$	$1$	$1$
$\dfrac{d^2y}{dx^2} + y = 0$	$2$	$1$
$\left(\dfrac{d^3y}{dx^3}\right)^3 + x\left(\dfrac{d^2y}{dx^2}\right)^2 = 0$	$3$	$3$
$\dfrac{d^3y}{dx^3} + 2\left(\dfrac{d^2y}{dx^2}\right)^2 - \dfrac{dy}{dx} + y = 0$	$3$	$1$
$\left(\dfrac{dy}{dx}\right)^2 + 2\left(\dfrac{dy}{dx}\right)^2 - \sin y = 0$	$1$	$2$
$\dfrac{dy}{dx} + \sin\!\left(\dfrac{dy}{dx}\right) = 0$	$1$	not defined

General and particular solutions, and solution curves

A number satisfies an algebraic equation like $x^2 + 1 = 0$ ; a function satisfies a differential equation. Substitute the function and its derivatives into the equation and, if the left side reduces to the right side identically, the function is a solution, and the curve $y = \varphi(x)$ it traces is called a solution curve or integral curve. Consider $\dfrac{d^2y}{dx^2} + y = 0$ : the function $y = a\sin(x+b)$ , for any real $a, b$ , satisfies it — substitute and the two derivative terms cancel exactly. Because this family carries two arbitrary constants (matching the order, two), it is the general solution, also called the primitive. Fix $a = 2$ and $b = \pi/4$ and you obtain $y = 2\sin\!\left(x + \pi/4\right)$ , one single curve with no free constants left — a particular solution.

\text{General solution: as many independent arbitrary constants as the order.} \qquad (3)

where: "independent" matters — two constants that always appear as a single combination (like $C = c_1 + c_2$ ) count as one effective constant, not two, and would signal that the family is really only a first order general solution in disguise. A particular solution is any member of the family obtained by assigning specific numerical values to every arbitrary constant, typically fixed by an initial condition such as " $y = 1$ when $x = 0$ " (an initial value problem) or by requiring the curve to pass through a named point.

Figure

A family of integral curves y equals x squared plus log of x plus C for three values of the constant C, with the particular solution through the point one comma one highlighted in red

Variables separable differential equations

A first order, first degree equation is separable when the right-hand side factors cleanly into a function of $x$ alone times a function of $y$ alone.

\frac{dy}{dx} = g(x)\,h(y) \quad \Longrightarrow \quad \frac{dy}{h(y)} = g(x)\,dx \quad \Longrightarrow \quad \int \frac{dy}{h(y)} = \int g(x)\,dx + C. \qquad (4)

where: the middle step is only valid when $h(y) \ne 0$ ; any constant value of $y$ that makes $h(y) = 0$ is itself a solution (often lost silently if you divide by $h(y)$ without checking), and $H(y) = G(x) + C$ is the general solution once both antiderivatives are found. This is the method behind NCERT's Examples 4 through 9 — from a purely algebraic separation such as $(2-y)\,dy = (x+1)\,dx$ to modelling continuously compounding interest, where $\dfrac{dP}{dt} = \dfrac{5}{100}P$ separates into $\dfrac{dP}{P} = \dfrac{dt}{20}$ and integrates to the familiar exponential growth law $P = Ce^{t/20}$ .

Homogeneous differential equations

A function $F(x,y)$ is homogeneous of degree $n$ if scaling both inputs by the same nonzero constant $\lambda$ scales the output by $\lambda^n$ : $F(\lambda x, \lambda y) = \lambda^n F(x,y)$ . A differential equation $\dfrac{dy}{dx} = F(x,y)$ is called homogeneous precisely when $F$ is homogeneous of degree zero — equivalently, when $F(x,y)$ can be written as a function of the single ratio $y/x$ alone.

\frac{dy}{dx} = F(x,y) = g\!\left(\frac{y}{x}\right). \qquad (5)

where: degree-zero homogeneity is the test — check $F(\lambda x, \lambda y) = \lambda^0 F(x,y) = F(x,y)$ , unchanged by the scaling. The substitution $y = vx$ , developed in full in the first derivation below, converts equation (5) into a variables separable equation in $v$ and $x$ ; when the equation is more naturally written as $\dfrac{dx}{dy} = h(x/y)$ , the mirror substitution $x = vy$ is used instead.

Figure

Rays from the origin at five different slopes, illustrating that the ratio y over x is constant along each ray, which is why the substitution y equals v x separates a homogeneous equation

Linear differential equations and the integrating factor

An equation is a first order linear differential equation when the dependent variable and its first derivative appear only to the first power, with no product of $y$ and $\dfrac{dy}{dx}$ between them.

\frac{dy}{dx} + Py = Q, \qquad P, Q \text{ functions of } x \text{ only (or constants).} \qquad (6)

where: the mirror form $\dfrac{dx}{dy} + P_1 x = Q_1$ , with $P_1, Q_1$ functions of $y$ alone, is used whenever the equation is more naturally linear in $x$ than in $y$ — recognising this alternate form, as in NCERT Example 22, is a favourite competitive-exam trick. The integrating factor, derived in full below, is $\text{I.F.} = e^{\int P\,dx}$ , and multiplying equation (6) through by it collapses the left side to $\dfrac{d}{dx}\!\left(y \cdot \text{I.F.}\right)$ , so the solution follows by direct integration:

y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\, dx + C. \qquad (7)

where: you must always divide through so the coefficient of $\dfrac{dy}{dx}$ is exactly $1$ before reading off $P$ and $Q$ — forgetting this step (as in $x\dfrac{dy}{dx} + 2y = x^2$ , where $P = 2/x$ only after dividing by $x$ ) is the single most common arithmetic slip in this section.

Derivations

Derivation AReducing a homogeneous differential equation to variables separable form

The goal is to prove that the substitution $y = vx$ always converts a homogeneous differential equation $\dfrac{dy}{dx} = F(x,y)$ into an equation whose variables separate cleanly, deriving the general working formula NCERT quotes as its equation (6) in Section 9.4.2. This single substitution is the master technique behind every homogeneous-type example in the chapter, from the algebraic $(x-y)\dfrac{dy}{dx} = x+2y$ to the trigonometric $x\cos(y/x)\dfrac{dy}{dx} = y\cos(y/x) + x$ , so proving it once in general saves re-deriving it for every new right-hand side.

Step 1. Start from $\dfrac{dy}{dx} = F(x,y)$ where $F$ is homogeneous of degree zero, meaning $F(\lambda x, \lambda y) = F(x,y)$ for every nonzero $\lambda$ . Choosing $\lambda = 1/x$ (valid whenever $x \ne 0$ ) gives $F(x,y) = F\!\left(1, \dfrac{y}{x}\right)$ , a function of the single ratio $y/x$ alone; call it $g(y/x)$ . The entire two-variable right-hand side has collapsed to a function of one variable, $v = y/x$ .

Step 2. Substitute $y = vx$ , treating $v$ as a new unknown function of $x$ . Differentiate the product $vx$ with respect to $x$ using the product rule: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ , since $\dfrac{d}{dx}(vx) = v \cdot 1 + x \cdot \dfrac{dv}{dx}$ .

Step 3. Substitute both $y = vx$ and $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ into the original equation, which becomes $v + x\dfrac{dv}{dx} = g(v)$ , since the right-hand side only ever depended on $y/x = v$ in the first place.

Step 4. Isolate the derivative term: $x\dfrac{dv}{dx} = g(v) - v$ . This is now an equation in exactly two variables, $v$ and $x$ , and — crucially — the right side is a pure function of $v$ while the coefficient on the left is a pure function of $x$ , which is exactly the separable form of equation (4).

Step 5. Separate and integrate, provided $g(v) \ne v$ : $\displaystyle\int \frac{dv}{g(v) - v} = \int \frac{dx}{x} = \ln|x| + C$ . Replacing $v$ by $y/x$ throughout recovers the general solution of the original homogeneous equation in $x$ and $y$ .

Step 6. Note the symmetric case. When the equation is naturally written $\dfrac{dx}{dy} = h(x/y)$ instead, the identical argument with the roles of $x$ and $y$ exchanged justifies the substitution $x = vy$ , giving $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ and the matching separable equation $\displaystyle\int \frac{dv}{h(v)-v} = \int \frac{dy}{y}$ . Choosing between $y=vx$ and $x=vy$ is simply a matter of which derivative, $dy/dx$ or $dx/dy$ , the equation was given in.

Go deeper · Derivation B · 2 minDeriving the integrating factor for a first order linear differential equation

Derivation BDeriving the integrating factor for a first order linear differential equation

The goal is to construct, from scratch, the function that NCERT calls the integrating factor and to prove that multiplying the standard linear equation $\dfrac{dy}{dx} + Py = Q$ by it turns the left-hand side into the derivative of a single product — the one move that makes the entire equation directly integrable. This is the most heavily used result in the chapter, underlying every linear-type worked example including the "linear in $x$ " trick tested at competitive level.

Step 1. Suppose some unknown positive function $g(x)$ , multiplied through the equation, will make life easier. Multiplying both sides of $\dfrac{dy}{dx} + Py = Q$ by $g(x)$ gives $g(x)\dfrac{dy}{dx} + P\,g(x)\,y = Q\,g(x)$ .

Step 2. Demand that the left-hand side be exactly the derivative of the product $y \cdot g(x)$ . By the product rule, $\dfrac{d}{dx}\!\left[y\,g(x)\right] = g(x)\dfrac{dy}{dx} + y\,g'(x)$ . Comparing this term by term against the left side of Step 1 forces $P\,g(x) = g'(x)$ — the coefficient of $y$ must match the derivative of $g$ itself.

Step 3. This condition on $g$ is itself a variables separable equation: $\dfrac{g'(x)}{g(x)} = P(x)$ . Integrating both sides with respect to $x$ , and noting the left side is exactly $\dfrac{d}{dx}\left[\ln g(x)\right]$ by the chain rule, gives $\ln g(x) = \displaystyle\int P\,dx$ (taking $g(x) > 0$ so no modulus is needed).

Step 4. Exponentiate both sides to solve for $g(x)$ explicitly: $g(x) = e^{\int P\,dx}$ . This function is defined to be the integrating factor, written $\text{I.F.} = e^{\int P\,dx}$ , and it was constructed to be exactly the multiplier that turns the left side of the original equation into an exact derivative.

Step 5. With $g(x) = \text{I.F.}$ substituted back, the original equation reads $\dfrac{d}{dx}\!\left[y \cdot \text{I.F.}\right] = Q \cdot \text{I.F.}$ , by construction from Step 2. Integrate both sides directly with respect to $x$ : $y \cdot \text{I.F.} = \displaystyle\int Q \cdot \text{I.F.}\, dx + C$ , which is the general solution formula quoted as equation (7) above.

Step 6. Check that the arbitrary constant produced while computing $\int P\,dx$ in Step 3 never matters. Adding any constant $k$ there multiplies the integrating factor by the fixed nonzero number $e^k$ ; this same factor multiplies every term of the equation in Step 5 uniformly and cancels out, leaving the same general solution. This is why you should always take the simplest antiderivative of $P$ when building the integrating factor — any other choice works too, but adds no new information.

Go deeper · Derivation C · 1 minVerifying that a two-parameter family solves a second order linear differential equation

Derivation CVerifying that a two-parameter family solves a second order linear differential equation

The goal is to verify, by direct and systematic differentiation, that $y = e^{ax}\!\left(c_1\cos bx + c_2 \sin bx\right)$ — a function carrying two independent arbitrary constants $c_1, c_2$ — satisfies the second order linear equation $\dfrac{d^2y}{dx^2} - 2a\dfrac{dy}{dx} + (a^2+b^2)y = 0$ for every choice of $c_1$ and $c_2$ . Beyond confirming NCERT's own miscellaneous Example 19, this derivation demonstrates the general technique for every "verify that the given function is a solution" question in Exercise 9.2, and it illustrates concretely why a second order equation's general solution needs exactly two arbitrary constants, matching its order.

Step 1. Write $y = e^{ax}\left[c_1\cos bx + c_2 \sin bx\right]$ and differentiate once using the product rule, treating $e^{ax}$ and the bracketed trigonometric sum as the two factors. The trigonometric part differentiates to $-bc_1\sin bx + bc_2\cos bx$ , so $\dfrac{dy}{dx} = e^{ax}\left[-bc_1\sin bx + bc_2\cos bx\right] + a\,e^{ax}\left[c_1\cos bx + c_2\sin bx\right]$ , which collects into $\dfrac{dy}{dx} = e^{ax}\left[(ac_1 + bc_2)\cos bx + (ac_2 - bc_1)\sin bx\right]$ .

Step 2. Differentiate this result once more, again by the product rule, to obtain $\dfrac{d^2y}{dx^2}$ . The trigonometric bracket in Step 1 differentiates to $-b(ac_1+bc_2)\sin bx + b(ac_2-bc_1)\cos bx$ , and adding the $a$ -multiplied copy of the Step 1 result and collecting cosine and sine terms separately gives $\dfrac{d^2y}{dx^2} = e^{ax}\left[(a^2c_1 + 2abc_2 - b^2c_1)\cos bx + (a^2c_2 - 2abc_1 - b^2c_2)\sin bx\right]$ .

Step 3. Substitute $y$ , $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ from Steps 1 and 2 into the left side of the differential equation, $\dfrac{d^2y}{dx^2} - 2a\dfrac{dy}{dx} + (a^2+b^2)y$ , keeping every $\cos bx$ term together and every $\sin bx$ term together (both are multiplied throughout by the common factor $e^{ax}$ , which never vanishes and can be factored out).

Step 4. Collect the coefficient of $\cos bx$ : $\left(a^2c_1 + 2abc_2 - b^2c_1\right) - 2a\left(ac_1+bc_2\right) + \left(a^2+b^2\right)c_1$ . Expanding the middle term gives $-2a^2c_1 - 2abc_2$ , and adding all three groups: $a^2c_1 + 2abc_2 - b^2c_1 - 2a^2c_1 - 2abc_2 + a^2c_1 + b^2c_1 = 0$ — every term cancels exactly.

Step 5. Collect the coefficient of $\sin bx$ the same way: $\left(a^2c_2 - 2abc_1 - b^2c_2\right) - 2a\left(ac_2-bc_1\right) + \left(a^2+b^2\right)c_2$ . Expanding the middle term gives $-2a^2c_2+2abc_1$ , and adding: $a^2c_2 - 2abc_1 - b^2c_2 - 2a^2c_2 + 2abc_1 + a^2c_2 + b^2c_2 = 0$ — this also cancels exactly.

Step 6. Since both coefficients vanish identically, the entire left side reduces to $e^{ax}[0\cdot\cos bx + 0\cdot\sin bx] = 0$ , matching the right side for every value of $c_1$ and $c_2$ — not for special values, but as an algebraic identity. This confirms $y = e^{ax}(c_1\cos bx + c_2\sin bx)$ is a genuine general solution of the order-two equation, carrying exactly two independent arbitrary constants, consistent with the count-equals-order rule.

Worked examples

Board level - Solving an initial value problem by variables separable, NCERT Example 6

Find the particular solution of $\dfrac{dy}{dx} = -4xy^2$ , given that $y = 1$ when $x = 0$ .

Step 1. Confirm the equation is separable and, for $y \ne 0$ , separate the variables by dividing through by $y^2$ and multiplying by $dx$ : $\dfrac{dy}{y^2} = -4x\,dx$ .

Step 2. Integrate both sides independently: $\displaystyle\int \frac{dy}{y^2} = -4\int x\,dx$ gives $-\dfrac{1}{y} = -2x^2 + C$ , which rearranges to $\dfrac{1}{y} = 2x^2 - C$ , i.e. $y = \dfrac{1}{2x^2 - C}$ .

Step 3. Apply the initial condition $y = 1$ at $x = 0$ : $1 = \dfrac{1}{0 - C}$ , so $-C = 1$ , giving $C = -1$ .

Step 4. Substitute $C = -1$ back into the general solution: $y = \dfrac{1}{2x^2 - (-1)} = \dfrac{1}{2x^2 + 1}$ . Answer: $y = \dfrac{1}{2x^2+1}$ .

Board level - Solving a first order linear differential equation, NCERT Example 15

Find the general solution of $x\dfrac{dy}{dx} + 2y = x^2$ , for $x \ne 0$ .

Step 1. Divide throughout by $x$ so the coefficient of $\dfrac{dy}{dx}$ becomes exactly $1$ : $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$ . This is now in the standard linear form with $P = 2/x$ and $Q = x$ .

Step 2. Compute the integrating factor: $\text{I.F.} = e^{\int (2/x)\,dx} = e^{2\ln x} = e^{\ln x^2} = x^2$ .

Step 3. Multiply the standard-form equation through by $x^2$ and integrate, using the fact that the left side collapses to $\dfrac{d}{dx}(y\,x^2)$ : $y\,x^2 = \displaystyle\int (x)(x^2)\,dx = \int x^3\,dx = \dfrac{x^4}{4} + C$ .

Step 4. Solve for $y$ by dividing through by $x^2$ : $y = \dfrac{x^2}{4} + Cx^{-2}$ . Answer: $y = \dfrac{x^2}{4} + Cx^{-2}$ .

Competitive level (JEE / NEET) - Recognising a linear-in-x equation, NCERT Example 22

Solve $\left(\tan^{-1}y - x\right)dy = \left(1+y^2\right)dx$ .

Step 1. This equation is not separable and not homogeneous in the usual $y/x$ sense — the trap is spending time forcing it into one of those two moulds. Instead, rewrite it as a relation between $\dfrac{dx}{dy}$ and $x$ : dividing both sides by $\left(1+y^2\right)dy$ gives $\dfrac{dx}{dy} = \dfrac{\tan^{-1}y - x}{1+y^2}$ , which rearranges to $\dfrac{dx}{dy} + \dfrac{x}{1+y^2} = \dfrac{\tan^{-1}y}{1+y^2}$ — linear, but in $x$ as a function of $y$ , with $P_1 = \dfrac{1}{1+y^2}$ and $Q_1 = \dfrac{\tan^{-1}y}{1+y^2}$ .

Step 2. Compute the integrating factor with respect to $y$ : $\text{I.F.} = e^{\int P_1\,dy} = e^{\int dy/(1+y^2)} = e^{\tan^{-1}y}$ .

Step 3. Multiply through and integrate: $x\,e^{\tan^{-1}y} = \displaystyle\int \frac{\tan^{-1}y}{1+y^2}\,e^{\tan^{-1}y}\,dy + C$ . Substitute $t = \tan^{-1}y$ , so $dt = \dfrac{dy}{1+y^2}$ , turning the integral into $\displaystyle\int t\,e^t\,dt = t\,e^t - e^t = e^t(t-1)$ by integration by parts, i.e. $e^{\tan^{-1}y}\left(\tan^{-1}y - 1\right)$ .

Step 4. Substitute back and divide through by $e^{\tan^{-1}y}$ : $x\,e^{\tan^{-1}y} = e^{\tan^{-1}y}\left(\tan^{-1}y - 1\right) + C$ , so $x = \left(\tan^{-1}y - 1\right) + Ce^{-\tan^{-1}y}$ . Answer: $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$ . Timing note: under three minutes once the linear-in-x recognition is made. The entire time cost sits in Step 1 — spotting that the equation is unworkable as $dy/dx$ but clean as $dx/dy$ linear. Never spend more than thirty seconds trying separation or the $y=vx$ substitution before checking whether flipping the derivative reveals a hidden linear form; this exact trap appears whenever an inverse trigonometric function of $y$ sits multiplied against $x$ .

Common traps

Confusing order with degree. Order counts how many times you differentiated; degree is the power of the highest derivative, and only exists when the equation is a polynomial in its derivatives — an equation with $\sin(y')$ , $e^{y''}$ , or $y'$ under a square root has an order but no degree at all.
Forgetting the arbitrary constant while integrating. Writing $H(y) = G(x)$ without the $+C$ silently throws away the general solution and returns only one particular member of the family.
Losing solutions when dividing by $h(y)$ or $g(x)$ during separation. Any constant value of $y$ that makes $h(y) = 0$ is itself a solution and can vanish from view the moment you divide it away without checking for it separately.
Testing homogeneity by eyeballing rather than by the $F(\lambda x, \lambda y) = \lambda^0 F(x,y)$ check. A right side that merely looks balanced in $x$ and $y$ is not automatically degree zero; always verify the scaling condition or rewrite the ratio as a function of $y/x$ explicitly before proceeding.
Using $y = vx$ when $x = vy$ is the natural substitution, or vice versa. Match the substitution to the derivative the equation was actually given in: a $\dfrac{dy}{dx}$ -form homogeneous equation takes $y=vx$ ; a $\dfrac{dx}{dy}$ -form equation takes $x=vy$ .
Reading off $P$ and $Q$ before the coefficient of the derivative is $1$ . In $x\dfrac{dy}{dx} + 2y = x^2$ , $P$ is not $2$ ; you must divide through by $x$ first to get $P = 2/x$ , exactly the slip in NCERT Example 15.
Sign errors inside the exponent of the integrating factor. A negative or fractional $P$ , such as $P = -1/y$ in Example 16's $\dfrac{dx}{dy} - \dfrac{x}{y} = 2y$ , is easy to mishandle; always simplify $\int P\,dx$ fully — using $\log$ laws where needed — before exponentiating.
Assuming every differential equation in this chapter can be solved by one universal recipe. Only three specific first order, first degree forms are covered; recognising which of the three (or, as in Example 22, a form that becomes one of the three after flipping the derivative) an equation belongs to is itself half the problem.
Forgetting the constant-count check. The number of independent arbitrary constants in a general solution always equals the order of the equation — a quick sanity check that catches an under- or over-integrated answer immediately, and is directly tested in Exercise 9.2 and the Miscellaneous Exercise.

Common trapOrder and degree are not the same number, and degree is not always defined

The single costliest habit in this chapter is treating "order" and "degree" as interchangeable, or assuming every differential equation has both. Order is always well defined — simply the highest derivative present — but degree exists only when the equation is a polynomial equation in its derivatives, with every derivative appearing to a non-negative integer power and never inside a transcendental function, a fraction with a derivative in the denominator, or under a radical. An equation such as $\dfrac{dy}{dx} + \sin\!\left(\dfrac{dy}{dx}\right) = 0$ has order one but degree not defined — not degree zero, not degree one by default. Examiners build entire questions (see the NCERT Exercise 9.1 objective questions) around students who reflexively write down a numerical degree without first checking whether the polynomial condition holds at all.

You should be able to…

State the order and degree of a given differential equation, and recognise when degree is not defined because the equation is not polynomial in its derivatives.
Explain, using the count of arbitrary constants, the difference between the general solution and a particular solution of a differential equation.
Verify by direct differentiation and substitution that a given function (explicit or implicit) satisfies a stated differential equation.
Recognise whether a first order first degree equation is variables separable, homogeneous, or linear, from its algebraic form alone.
Solve a variables separable equation, correctly handling the constant of integration and any solutions lost by dividing through $h(y)$ .
Show that a given $F(x,y)$ is homogeneous of degree zero and solve the resulting equation using the substitution $y=vx$ (or $x=vy$ ).
Reduce a first order linear differential equation to standard form, compute its integrating factor, and write down the general solution.
Recognise the "linear in $x$ " trick when an equation resists separation and homogeneity in its given $dy/dx$ form.
Find a particular solution satisfying a given initial condition, and find the equation of a curve passing through a specified point.

Ask the tutor: Want to work through a few more mixed-type equations — where you have to decide separable, homogeneous, or linear before solving — until the recognition step becomes instant?

What is actually the difference between order and degree, in plain terms

Order tells you how many times the function has been differentiated — it is simply the label on the highest derivative present, and every ordinary differential equation has one. Degree tells you the power that highest derivative is raised to, but only when the whole equation can be written as a polynomial in its derivatives with non-negative integer powers. If a derivative sits inside a sine, an exponential, a logarithm, or under a square root, the equation is not a polynomial in its derivatives, and degree is simply not defined for it — order can still be found perfectly well in that same equation.

Why does a second order equation need two arbitrary constants but a first order equation only needs one

Each order of differentiation you undo by integrating introduces exactly one fresh constant of integration, because indefinite integration is only ever determined up to an additive constant. Solving a first order equation involves one integration, so one constant appears. A second order equation, like $\dfrac{d^2y}{dx^2}+y=0$ , effectively needs two integrations to get back to $y$ itself, so two independent constants survive, as shown explicitly in the third derivation above for $y=e^{ax}(c_1\cos bx+c_2\sin bx)$ . This is why "count the constants, compare to the order" is such a reliable self-check.

How do I decide whether an equation is separable, homogeneous, or linear before I start solving it

Work through three quick checks in order. First, try to physically separate the variables — can every $y$ -term be herded to one side and every $x$ -term to the other, even after some algebraic rearrangement? If yes, it is separable. If not, check whether the right side of $dy/dx = F(x,y)$ is unchanged when you replace $x$ by $\lambda x$ and $y$ by $\lambda y$ — if so, it is homogeneous and $y=vx$ applies. If neither works, look for the linear pattern: does $y$ (or $x$ ) and its own first derivative appear only to the first power, with no $y\cdot dy/dx$ product? If so, divide to standard form and read off $P$ and $Q$ . Example 22 shows the fourth possibility worth remembering: an equation that fails all three tests in $dy/dx$ form can still be linear once you rewrite it in terms of $dx/dy$ instead.

Do I always have to use y equals v x for homogeneous equations, or can I use x equals v y

Either substitution works on a genuinely homogeneous equation, but one is usually far less messy than the other, and NCERT's own convention is to match the substitution to the form the derivative is given in. If the equation is written as $dy/dx = g(y/x)$ , use $y=vx$ , since $y$ is already the dependent variable being solved for. If it is more naturally $dx/dy = h(x/y)$ , use $x=vy$ instead. Using the "wrong" one still gets you to a correct answer eventually, but usually after considerably messier algebra, since you would be fighting the natural direction of the equation the whole way.

Why does multiplying by the integrating factor suddenly make the equation solvable

Because the integrating factor is not chosen at random — it is constructed, in the second derivation above, to be exactly the function that makes the left side of $dy/dx + Py = Q$ collapse into the derivative of a single product, $\dfrac{d}{dx}(y \cdot \text{I.F.})$ . Once the left side is a single exact derivative, the entire equation becomes "the derivative of something equals a known function of $x$ ," which integration undoes directly. The apparent magic of the integrating factor is really just running the product rule in reverse, engineered on purpose by solving $g'(x)=Pg(x)$ for the right multiplier in the first place.