JEE Main & Advanced

Differential Equations

Integral Calculus: Differential Equations

1
Module 1

Foundations of Differential Equations

Order, Degree, and Formation of EquationsTopic 1

A differential equation (DE) is an equation containing an independent variable, a dependent variable, and the derivatives of the dependent variable with respect to the independent variable. The classification and behavior of a DE depend on its order and degree:
  • Order: The order of a differential equation is the order of the highest-order derivative appearing in the equation. It is always a positive integer and is unconditionally defined for every DE.
  • Degree: The degree of a differential equation is the power or exponent of the highest-order derivative, provided the equation can be expressed as a polynomial equation in terms of its derivatives. If the derivatives are embedded inside transcendental functions (such as $\sin(y')$, $\ln(y'')$, or $e^{y'}$), the degree of the equation is strictly undefined.
To form a differential equation from a given family of curves, differentiate the geometric equation successively with respect to the independent variable $x$. If the curve equation contains $n$ independent arbitrary constants, you must differentiate the equation exactly $n$ times to generate a system of equations. Eliminating the arbitrary constants from this system yields a differential equation of order $n$.
Worked Examples
1

Determine the order and degree of the following differential equation: \[ y = x \frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \]

Show solution

The highest-order derivative in the equation is the first derivative, $\frac{dy}{dx}$. Therefore, the order of the equation is 1. To find the degree, we must first remove the radical sign to express the equation as a polynomial in terms of its derivatives. Rearrange the terms: \[ y - x \frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \] Square both sides of the equation to eliminate the square root: \[ \left(y - x \frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \] Expand the left side: \[ y^2 - 2xy \frac{dy}{dx} + x^2 \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{dy}{dx}\right)^2 \] Gather all terms onto one side: \[ (x^2 - 1)\left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + (y^2 - 1) = 0 \] The equation is now expressed as a polynomial in terms of its derivatives. The exponent of the highest-order derivative ($\frac{dy}{dx}$) is 2. Therefore, the degree of the equation is 2. Final Answer: Order = 1, Degree = 2.

✎ Self-Check — 5 questions0 / 5
Q1.Determine the order and degree of the differential equation $\frac{d^2y}{dx^2} + \cos\left(\frac{dy}{dx}\right) = 0$:
Q2.Find the order of the differential equation corresponding to the family of curves $y = c_1 e^{x+c_2} + c_3 e^{-x}$:
Q3.Form the differential equation for the family of circles passing through the origin and having their centers on the x-axis:
Q4.Find the degree of the differential equation $\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{5/3} = \frac{d^2y}{dx^2}$:
Q5.The differential equation representing the family of parabolas $y^2 = 4ax$ is:
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Module 2

Solution Methodologies for First-Order Equations

Separation of Variables and SubstitutionsTopic 1

The most straightforward method for solving a first-order differential equation is the Variable Separable Form. If the equation can be factored such that the derivative $\frac{dy}{dx}$ splits into a product of a function of $x$ and a function of $y$, you can group all terms containing $y$ with the differential $dy$, and all terms containing $x$ with the differential $dx$: \[ \frac{dy}{dx} = f(x)g(y) \implies \frac{1}{g(y)} \, dy = f(x) \, dx \] Integrating both sides directly yields the general solution: $\int \frac{1}{g(y)} \, dy = \int f(x) \, dx + C$.

When a differential equation is not immediately separable, it can often be reduced to variable separable form using an algebraic substitution:
  • Linear Argument Forms: For equations of the type $\frac{dy}{dx} = f(ax + by + c)$, substitute $v = ax + by + c$. Differentiating this substitution with respect to $x$ yields $\frac{dv}{dx} = a + b\frac{dy}{dx}$, which transforms the original equation into a separable equation in terms of $v$ and $x$.
  • Homogeneous Forms: A differential equation is homogeneous if it can be written in the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$. To solve this form, substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. This transformation simplifies the equation into a separable form in terms of $v$ and $x$: \[ v + x\frac{dv}{dx} = F(v) \implies x\frac{dv}{dx} = F(v) - v \implies \frac{1}{F(v) - v} \, dv = \frac{1}{x} \, dx \]
Worked Examples
1

Solve the following homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.

Show solution

Divide the numerator and denominator by $x^2$ to express the right side as a function of the ratio $\frac{y}{x}$: \[ \frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{2\left(\frac{y}{x}\right)} \] Apply the standard homogeneous substitution: $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. \[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v} \] Subtract $v$ from both sides to isolate the derivative term: \[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \] Separate the variables by moving all terms containing $v$ to the left side and all terms containing $x$ to the right side: \[ \frac{2v}{1 - v^2} \, dv = \frac{1}{x} \, dx \] Integrate both sides. Use substitution on the left side ($u = 1 - v^2 \implies du = -2v \, dv$): \[ -\ln|1 - v^2| = \ln|x| + C_1 \implies \ln|x| + \ln|1 - v^2| = -C_1 \] Combine the logarithmic terms: \[ \ln|x(1 - v^2)| = -C_1 \implies x(1 - v^2) = e^{-C_1} = C \] Substitute the original variable ratio $v = \frac{y}{x}$ back into the equation: \[ x\left(1 - \frac{y^2}{x^2}\right) = C \implies x\left(\frac{x^2 - y^2}{x^2}\right) = C \implies \frac{x^2 - y^2}{x} = C \implies x^2 - y^2 = Cx \] Final Answer: $x^2 - y^2 = Cx$.

✎ Self-Check — 5 questions0 / 5
Q1.Solve the separable differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$:
Q2.Solve the differential equation $\frac{dy}{dx} = (4x + y + 1)^2$ by using an appropriate linear substitution:
Q3.The general solution of the homogeneous equation $\frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right)$ is:
Q4.Solve the separable equation $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$:
Q5.To reduce the differential equation $\frac{dy}{dx} = \frac{x-y+3}{2x-2y+5}$ to variable separable form, the most effective substitution is:

Linear First-Order Equations, Bernoulli's Form, and ExactnessTopic 2

A first-order differential equation is linear if it can be written in the standard form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where $P(x)$ and $Q(x)$ are continuous functions of $x$. To solve this equation, multiply all terms by an Integrating Factor (IF), denoted by $\mu(x) = e^{\int P(x) \, dx}$. Multiplying by the integrating factor transforms the left side into the derivative of a product, allowing you to integrate directly: \[ \frac{d}{dx}\left[ y \cdot e^{\int P(x) \, dx} \right] = Q(x) \cdot e^{\int P(x) \, dx} \implies y \cdot \mu(x) = \int Q(x) \mu(x) \, dx + C \]

Two advanced variations of this first-order form are frequently tested in competitive exams:
  • Bernoulli's Equation: A non-linear equation containing a power of the dependent variable on the right side: $\frac{dy}{dx} + P(x)y = Q(x)y^n$. To linearize this equation, divide all terms by $y^n$ and substitute $v = y^{1-n}$. This transforms the expression into a standard linear equation in terms of $v$ and $x$.
  • Exact Differential Equations: An equation expressed in differential form, $M(x,y) \, dx + N(x,y) \, dy = 0$, is exact if its components satisfy the partial derivative condition: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] The general solution is found by integrating $M$ with respect to $x$ (treating $y$ as a constant) and adding the integrals of the terms in $N$ that contain *only* $y$. If the exactness condition is not met, the equation can often be made exact by multiplying by a targeted integrating factor.
Worked Examples
1

Find the general solution of the following linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x^3 \quad (x > 0)$.

Show solution
The equation is already in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2}{x}$ and $Q(x) = x^3$.
  • Step 1: Calculate the Integrating Factor (IF)
    \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2 \]
  • Step 2: Apply the linear solution formula
    \[ y \cdot \mu(x) = \int Q(x)\mu(x) \, dx + C \] Substitute $\mu(x) = x^2$ and $Q(x) = x^3$ into the equation: \[ y \cdot x^2 = \int (x^3 \cdot x^2) \, dx + C = \int x^5 \, dx + C \] Integrate the right side using the standard power rule: \[ y \cdot x^2 = \frac{x^6}{6} + C \] Divide both sides by $x^2$ to isolate $y$ and find the general solution: \[ y = \frac{x^4}{6} + C x^{-2} \]
Final Answer: $y = \frac{x^4}{6} + \frac{C}{x^2}$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the integrating factor for the linear differential equation $\frac{dy}{dx} + y \tan x = \sec x$:
Q2.Solve the linear equation $x \frac{dy}{dx} - y = x^2$:
Q3.To linearize the Bernoulli differential equation $\frac{dy}{dx} + \frac{1}{x}y = x^2 y^3$, the most effective operational substitution is:
Q4.Which of the following differential equations is strictly exact?
Q5.Solve the exact differential equation $(3x^2 + 2xy) \, dx + (x^2 + y^2) \, dy = 0$:
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Module 3

Higher-Order Systems & Advanced Solution Forms

Second-Order Linear Equations with Constant CoefficientsTopic 1

A linear second-order differential equation with constant coefficients has the standard format: \[ a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = f(x) \] The complete general solution consists of two independent parts added together: $y(x) = y_c(x) + y_p(x)$, where $y_c(x)$ is the Complementary Function (CF) and $y_p(x)$ is the Particular Integral (PI).
  • Complementary Function ($y_c$): Found by setting the forcing function to zero ($f(x) = 0$) and solving the corresponding auxiliary quadratic equation $a m^2 + b m + c = 0$:
    1. Real and Distinct Roots ($m_1 \neq m_2$): $y_c = c_1 e^{m_1 x} + c_2 e^{m_2 x}$
    2. Real and Repeated Roots ($m_1 = m_2 = m$): $y_c = (c_1 + c_2 x)e^{m x}$
    3. Complex Conjugate Roots ($m = \alpha \pm i\beta$): $y_c = e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)$
  • Particular Integral ($y_p$): Accounts for the non-zero forcing function $f(x)$. It can be found using the method of undetermined coefficients (guessing a template solution based on the form of $f(x)$) or via the Variation of Parameters method for general functions.
Worked Examples
1

Find the complete general solution of the second-order differential equation: $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.

Show solution

The equation is a homogeneous second-order linear differential equation, so its solution consists entirely of the complementary function ($y = y_c$). Write down the corresponding auxiliary quadratic equation by replacing the derivatives with powers of $m$: \[ m^2 - 5m + 6 = 0 \] Factor the quadratic polynomial to find its roots: \[ (m - 2)(m - 3) = 0 \implies m_1 = 2, \quad m_2 = 3 \] The auxiliary equation has two real and distinct roots. Apply the corresponding solution template: \[ y(x) = c_1 e^{m_1 x} + c_2 e^{m_2 x} = c_1 e^{2x} + c_2 e^{3x} \] Final Answer: $y = c_1 e^{2x} + c_2 e^{3x}$.

✎ Self-Check — 5 questions0 / 5
Q1.Find the complementary function for the second-order linear equation $\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0$:
Q2.Find the auxiliary equation roots for the differential equation $y'' + 4y = 0$:
Q3.What is the guessed template form for the particular integral $y_p$ if the forcing function is $f(x) = e^{5x}$ and 5 is not a root of the auxiliary equation?
Q4.The general solution of the second-order differential equation $y'' + y' - 2y = 0$ is:
Q5.Find the general solution of the equation $y'' + 9y = 0$:

Clairaut's Equation, Geometric Isoclines, and Orthogonal TrajectoriesTopic 2

This subsection covers several advanced geometric and analytical forms often featured in JEE Advanced:
  • Clairaut's Equation: A first-order non-linear equation with the standard format $y = x p + f(p)$, where $p = \frac{dy}{dx}$. Its general solution is found by simply replacing the derivative term $p$ with an arbitrary constant $c$: $y = cx + f(c)$. This represents a family of straight lines. Differentiating with respect to $p$ yields the singular solution, which represents the envelope of these lines and contains no arbitrary constants.
  • Orthogonal Trajectories: An orthogonal trajectory is a curve that intersects every member of a given family of curves at a right angle ($90^\circ$). To find the orthogonal family:
    1. Form the differential equation for the given family of curves: $F(x, y, \frac{dy}{dx}) = 0$.
    2. Replace the original derivative term $\frac{dy}{dx}$ with its negative reciprocal, $-\frac{dx}{dy}$, to enforce perpendicularity.
    3. Solve the resulting new differential equation to find the orthogonal curves.
  • Slope Fields and Isoclines: Provide a method for visualizing solutions geometrically. An isocline for the equation $\frac{dy}{dx} = f(x,y)$ is a curve along which the slope of the solution path is constant ($f(x,y) = k$). Sketching these constant-slope line segments helps trace solution curves without solving the equation analytically.
Worked Examples
1

Find the general solution of the following Clairaut equation: $y = x p + p^2 \quad \left(p = \frac{dy}{dx}\right)$.

Show solution
The equation matches the standard Clairaut format $y = x p + f(p)$, where $f(p) = p^2$. To find the general solution, differentiate both sides of the equation with respect to $x$: \[ \frac{dy}{dx} = p = 1 \cdot p + x \frac{dp}{dx} + 2p \frac{dp}{dx} \] Cancel the $p$ terms on both sides: \[ 0 = (x + 2p)\frac{dp}{dx} \] This gives two distinct cases:
  • Case 1: $\frac{dp}{dx} = 0 \implies p = c$ (an arbitrary constant). Substituting $p = c$ back into the original equation yields the general solution family of straight lines: \[ y = cx + c^2 \]
  • Case 2: $x + 2p = 0 \implies p = -\frac{x}{2}$. Substituting this expression back into the original equation yields the singular solution envelope: \[ y = x\left(-\frac{x}{2}\right) + \left(-\frac{x}{2}\right)^2 = -\frac{x^2}{2} + \frac{x^2}{4} = -\frac{x^2}{4} \implies x^2 = -4y \]
Final Answer: The general solution is $y = cx + c^2$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the general solution of the Clairaut equation $y = x p + \frac{1}{p}$:
Q2.To find the orthogonal trajectories of a family of curves, the derivative term $\frac{dy}{dx}$ in its differential equation must be replaced with:
Q3.Find the orthogonal trajectories of the family of straight lines passing through the origin ($y = mx$):
Q4.Along an isocline curve for the first-order differential equation $\frac{dy}{dx} = f(x,y)$, the solution paths always exhibit:
Q5.Find the singular solution envelope for the Clairaut equation $y = x p + p^2$:
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Module 4

Physical Modeling & Applications

Physical Applications and Modeling SystemsTopic 1

Differential equations serve as the primary mathematical framework for modeling rate-dependent physical systems across science and engineering fields:
  • Exponential Growth and Decay: Modeled by the first-order separable equation $\frac{dN}{dt} = kN$. If $k > 0$, the system experiences exponential growth (e.g., unconstrained population models); if $k < 0$, it experiences exponential decay (e.g., radioactive substance half-life tracking).
  • Newton's Law of Cooling: States that the rate of change of the temperature $T$ of an object is directly proportional to the difference between its temperature and the surrounding ambient temperature $T_m$: \[ \frac{dT}{dt} = -k(T - T_m) \quad (\text{where } k > 0) \]
  • Mixing Problems: Track the chemical salt content $S(t)$ in a fluid tank over time. The rate of change is modeled using mass balance mechanics: \[ \frac{dS}{dt} = \text{Rate in} - \text{Rate out} = (\text{Inflow Concentration} \times \text{Inflow Rate}) - \left(\frac{S(t)}{\text{Volume}} \times \text{Outflow Rate}\right) \]
  • Projectile Motion with Air Resistance: Extends basic linear kinematics by adding a velocity-dependent drag force term ($F_{\text{drag}} = -cv$), transforming the system into a manageable first-order linear differential equation in terms of velocity.
Worked Examples
1

A radioactive substance decays at a rate proportional to its mass. If the initial mass is $100\text{ mg}$ and it half-decays to $50\text{ mg}$ in exactly $20\text{ hours}$, find the expression for the mass remaining at any time $t$.

Show solution

Let $M(t)$ represent the mass of the radioactive substance at time $t$. The decay process is modeled by the differential equation: \[ \frac{dM}{dt} = -kM \quad (\text{where } k > 0) \] Separate the variables and integrate both sides: \[ \frac{1}{M} \, dM = -k \, dt \implies \ln M = -kt + C_1 \implies M(t) = e^{-kt + C_1} = C e^{-kt} \] Apply the initial condition ($M(0) = 100$) to find the constant $C$: \[ 100 = C e^0 \implies C = 100 \implies M(t) = 100 e^{-kt} \] Now apply the second condition ($M(20) = 50$) to solve for the decay constant $k$: \[ 50 = 100 e^{-20k} \implies e^{-20k} = \frac{1}{2} \implies -20k = \ln\left(\frac{1}{2}\right) = -\ln 2 \implies k = \frac{\ln 2}{20} \] Substitute the expression for $k$ back into the mass equation to find the final tracking formula: \[ M(t) = 100 e^{-\left(\frac{\ln 2}{20}\right)t} = 100 \left(e^{\ln 2}\right)^{-t/20} = 100 (2)^{-t/20} \] Final Answer: $M(t) = 100 \cdot 2^{-t/20}$.

✎ Self-Check — 5 questions0 / 5
Q1.A population grows at a rate modeled by $\frac{dP}{dt} = 0.05P$. If the initial population is $P_0$, the population at time $t$ is:
Q2.According to Newton's Law of Cooling, if a cup of coffee at $90^\circ\text{C}$ is placed in a room kept at a constant temperature of $20^\circ\text{C}$, the initial differential setup tracking its temperature $T$ is:
Q3.A tank holds $100\text{ L}$ of brine solution. Pure water pumps in at $4\text{ L/min}$, and the mixed solution drains out at the same rate. The differential equation tracking the salt content $S(t)$ is:
Q4.The velocity of a falling object experiencing linear air resistance changes according to the equation $\frac{dv}{dt} = g - kv$. Its terminal velocity $v_{\text{terminal}}$ is achieved when $\frac{dv}{dt} = 0$, which evaluates to:
Q5.If a bacterial culture doubles in size every 4 hours under exponential growth conditions, its population size equation can be modeled as:

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