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Differential Equations

Equations involving derivatives and their solutions

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Reading time~9 min
Revision time~3 min
Last updated2026-08-18
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🎯 Key Points

  • Order = highest derivative present; Degree = power of that highest derivative (only defined when the equation is polynomial in its derivatives)
  • Linear first-order dy/dx+P(x)y=Q(x): integrating factor IF=e^∫P dx, solution y·IF=∫Q·IF dx - memorize this exact template, it solves the entire category
  • Number of arbitrary constants in the general solution = order of the DE; a particular solution fixes those constants via given initial conditions
  • To FORM a DE from a family of curves with n constants: differentiate n times, then eliminate the constants between the original relation and its derivatives
Family of Solution Curves: y = Ax²Each value of the constant A gives ONE particular curve from the family

The general solution y=Ax² represents an entire FAMILY of curves, one for each value of the arbitrary constant A; a particular solution (fixed by an initial condition) selects exactly one curve from this family.

Differential Equations

A differential equation (DE) contains derivatives. Solutions describe how quantities change over time or position.

Order and Degree

  • Order = highest derivative in the equation
  • Degree = power of the highest derivative (when polynomial)
  • Example: d²y/dx² + 3(dy/dx)² + y = 0 has order 2, degree 1

First Order DEs

  • Separable: dy/dx = f(x)g(y) → separate variables: dy/g(y) = f(x)dx, then integrate both sides
  • Linear: dy/dx + P(x)y = Q(x). Integrating factor = e^∫P dx. Solution: y × IF = ∫Q × IF dx
  • Homogeneous: Put y = vx, then dy/dx = v + x(dv/dx)

Second Order Linear DEs

For ay'' + by' + cy = 0:

  • Characteristic equation: am² + bm + c = 0
  • Two distinct real roots (m₁, m₂): y = Aem₁x + Bem₂x
  • Equal roots (m₁ = m₂ = m): y = (A + Bx)emx
  • Complex roots (m = α ± iβ): y = eαx(A cosβx + B sinβx)

Applications

  • Exponential growth/decay: dy/dt = ky → y = y₀ekt
  • Newton's law of cooling: dT/dt = -k(T - T₀)
  • Simple harmonic motion: d²x/dt² = -ω²x
  • Population models, radioactive decay, charging capacitors

Formation of a Differential Equation

A differential equation can be formed from a family of curves by eliminating the arbitrary constants. If a relation has n arbitrary constants, differentiate it n times and eliminate the constants between the original relation and its derivatives to get an equation of order n.

Worked example: Form the DE for the family y = Ax2 (A is an arbitrary constant).

Differentiating: dy/dx = 2Ax. From the original equation, A = y/x2, so dy/dx = 2x(y/x2) = 2y/x, which gives the DE x(dy/dx) = 2y, a first-order equation with the constant eliminated.

Worked Example: Variable Separable

Solve dy/dx = 2xy with the initial condition y(0) = 1.

Separate variables: dy/y = 2x dx. Integrating both sides: ln|y| = x2 + C, so y = A ex2 where A = eC.

Applying y(0) = 1: 1 = A e0 = A, so A = 1. The particular solution is y = ex2.

Worked Example: Linear Differential Equation

Solve dy/dx + y/x = x (for x > 0).

This is in the standard linear form dy/dx + P(x)y = Q(x) with P(x) = 1/x and Q(x) = x.

Integrating factor: IF = e^∫(1/x)dx = eln x = x.

Solution: y x IF = ∫Q x IF dx, so y x x = ∫x x x dx = ∫x2 dx = x3/3 + C.

Therefore y = x2/3 + C/x. This general solution contains one arbitrary constant, consistent with a first-order equation.

General Solution vs Particular Solution

  • A general solution contains as many independent arbitrary constants as the order of the differential equation.
  • A particular solution is obtained by assigning specific values to the constants, usually using given initial conditions (initial value problem) such as y(x0) = y0.
  • Geometrically, the general solution represents a family of curves, while a particular solution picks out exactly one curve from that family.

Verifying a Solution of a Differential Equation

A function y = φ(x) is a solution of a differential equation if it, together with its derivatives, satisfies the equation identically for all x in the domain.

  • Method: compute the required derivatives of the given function, substitute them into the differential equation, and check that the two sides become equal.
  • Example: verify that y = e-x + 1 is a solution of y'' + y' = 0. Here y' = -e-x and y'' = e-x, so y'' + y' = e-x - e-x = 0, which satisfies the equation.
  • A relation defining y implicitly can also be a solution; differentiate implicitly and substitute in the same way.

Homogeneous Differential Equations

A first-order equation dy/dx = F(x, y) is homogeneous if F can be written purely as a function of the ratio y/x, i.e. dy/dx = g(y/x). Equivalently it has the form dy/dx = f(x, y)/h(x, y) where f and h are homogeneous functions of the same degree.

  • Substitution: put y = vx, so dy/dx = v + x(dv/dx). This reduces the equation to a variables-separable form in v and x.
  • After separating and integrating, replace v by y/x to return to the original variables.
  • If the equation is more naturally a function of x/y, use the symmetric substitution x = vy with dx/dy = v + y(dv/dy) instead.
  • Example: for dy/dx = (x + y)/x = 1 + y/x, put y = vx to get v + x(dv/dx) = 1 + v, so x(dv/dx) = 1, giving dv = dx/x, hence v = ln|x| + C and y = x(ln|x| + C).

🚀 JEE Advanced Edge

Recognizing a "disguised" homogeneous or linear equation: An equation like dy/dx = (x+y)/(x-y) doesn't look separable, but dividing numerator and denominator by x reveals it depends only on the ratio y/x - the signature of a homogeneous equation, solved via y=vx. Many JEE differential equations require this kind of algebraic rearrangement BEFORE the standard method (separable, linear, or homogeneous) becomes visible.

Orthogonal trajectories: Two families of curves are orthogonal trajectories of each other if every curve in one family intersects every curve in the other family at right angles - found by taking the differential equation of one family, replacing dy/dx with -dx/dy (the perpendicular-slope condition), and solving the resulting DE to get the second family. This technique connects differential equations directly back to the perpendicular-slope idea from coordinate geometry.

Worked problem: Solve the differential equation dy/dx = y/x + tan(y/x) (a homogeneous equation in disguise). Approach: Substitute y=vx, so dy/dx=v+x(dv/dx). The equation becomes v+x(dv/dx) = v+tan(v), i.e. x(dv/dx)=tan(v). Separating: cot(v)dv = dx/x. Integrating: ln|sin v| = ln|x|+C, so sin(v) = Ax. Substituting back v=y/x: sin(y/x) = Ax is the general solution.

Basic Concepts of a Differential Equation

A differential equation is an equation involving the derivatives of an unknown (dependent) variable with respect to one or more independent variables.

  • Because it contains only one independent variable, the type studied in Class 12 is called an ordinary differential equation (ODE).
  • Examples include dy/dx = 2x, and the second-order equation d-squared-y/dx-squared + y = 0.
  • A solution of a differential equation is a function y = g(x) which, together with its derivatives, satisfies the equation identically.
  • Differential equations model growth, decay, motion, cooling and many physical situations where a rate of change is known.

Order and Degree of a Differential Equation

Order and degree classify a differential equation and decide how many arbitrary constants its general solution contains.

  • Order is the order of the highest derivative appearing in the equation (first order means dy/dx is the highest, second order means d-squared-y/dx-squared is the highest).
  • Degree is the power of the highest-order derivative, after the equation has been made a polynomial (free of radicals and fractions) in its derivatives.
  • Degree is defined only when the equation can be written as a polynomial in the derivatives; otherwise the degree is not defined.
  • The general solution of an equation of order n contains exactly n arbitrary constants.

The Variables Separable Method

A first-order equation of the form dy/dx = h(x) times g(y) can be solved by separating the variables so each side involves only one variable.

  • Rearrange to the form (1/g(y)) dy = h(x) dx, collecting all y-terms with dy and all x-terms with dx.
  • Integrate both sides; the result is the general solution, and a single arbitrary constant of integration is added.
  • An initial condition (a given value of y at a particular x) is used to evaluate that constant and obtain a particular solution.
  • This method applies only when the right side factors cleanly into a function of x times a function of y.

Linear Differential Equations and the Integrating Factor

A linear differential equation of the first order has the standard form dy/dx + P(x) times y = Q(x), where P and Q are functions of x (or constants).

  • Compute the integrating factor I.F. = e raised to the integral of P dx.
  • Multiply the whole equation by the I.F.; the left side becomes the derivative of (y times I.F.).
  • Integrate both sides to get y times (I.F.) = integral of Q times (I.F.) dx, plus an arbitrary constant c.
  • A similar form dx/dy + P(y) times x = Q(y) is solved the same way, treating x as the dependent variable.
2 Revise ~3 min before the exam

📐 Formula Sheet

  • Order: the highest derivative present  |  Degree: the power of that highest derivative (once radicals are cleared)
  • Variables separable: write as f(y)dy = g(x)dx, then integrate both sides
  • Homogeneous: dy/dx = f(y/x); substitute y = vx, so dy/dx = v + x·dv/dx
  • Linear (first order): dy/dx + Py = Q
  • Integrating factor: IF = e∫P dx
  • Solution of a linear ODE: y·(IF) = ∫Q·(IF)dx + C
  • General solution: contains as many arbitrary constants as the order
3 Practice apply it

✍️ Worked Examples

Example 1 - Variables separable
Q: Solve dy/dx = xy.
Step 1 - Separate variables: dy/y = x dx.
Step 2 - Integrate both sides: ln|y| = x²/2 + C.
Step 3 - Exponentiate: y = ex²/2 + C = A·ex²/2, where A = eC.
Answer: y = A·ex²/2. Note: the single arbitrary constant matches the first-order equation.

Example 2 - Linear first-order equation
Q: Solve dy/dx + 2y = e−x.
Step 1 - Identify: P = 2, Q = e−x.
Step 2 - Integrating factor: IF = e∫2 dx = e2x.
Step 3 - Apply y·IF = ∫Q·IF dx: y·e2x = ∫e−x·e2xdx = ∫exdx = ex + C.
Step 4 - Divide by e2x: y = e−x + C·e−2x.
Answer: y = e−x + Ce−2x.

Example 3 - Order and degree
Q: State the order and degree of (d²y/dx²)³ + 2(dy/dx)⁴ + y = 0.
Step 1 - Highest derivative present is d²y/dx², so the order is 2.
Step 2 - Its power is 3, and the equation is already polynomial in derivatives.
Step 3 - So the degree is 3.
Answer: order 2, degree 3. Trap: degree is the power of the highest-order derivative (here squared-term cubed = 3), not the largest power in the equation (the 4 on dy/dx is irrelevant).

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Frequently Asked Questions - Differential Equations

What are the key concepts in Differential Equations?
Equations involving derivatives and their solutions
Is Differential Equations important for JEE?
Yes. Differential Equations is part of the Mathematics Class 12 NCERT syllabus and is directly tested in JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Differential Equations questions on StudyHub?
Open StudyHub and select Mathematics → Differential Equations. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Mathematics Textbook - Chapter: Differential Equations
  2. CBSE Curriculum - Mathematics (Class 12)
  3. NTA JEE Main Official Syllabus - subject-wise topic list