Differential Equations revision notes

Key Concepts & Definitions

Differential Equation An equation involving an independent variable, a dependent variable, and derivatives of the dependent variable with respect to the independent variable. These equations are used to model applications in Physics, Chemistry, Biology, Economics, and more.

Ordinary Differential Equation (ODE) A differential equation involving derivatives of the dependent variable with respect to only one independent variable. Throughout these notes, the term 'differential equation' refers exclusively to ordinary differential equations.

Order of a Differential Equation The order of a differential equation is the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the equation. Order is always a positive integer. → [JEE TIP] The order is intrinsic to the equation and cannot be altered by algebraic manipulations.

Degree of a Differential Equation To study the degree, the differential equation must be a polynomial equation in its derivatives (i.e., $y', y'', y'''$, etc.). By degree, we mean the highest power (positive integral index) of the highest order derivative involved in the equation. If the equation cannot be written as a polynomial in its derivatives (e.g., involves $\sin(y'), e^{y'}, \ln(y')$), the degree is not defined. Degree, when defined, is always a positive integer.

Solutions of a Differential Equation A solution is a function $\phi$ (free from derivatives) that satisfies the differential equation; when substituted into the equation, the L.H.S. equals the R.H.S. The curve $y = \phi(x)$ is called the solution curve or integral curve.

General Solution (Primitive):

The solution which contains as many arbitrary constants (parameters) as the order of the differential equation.

Particular Solution:

A solution obtained from the general solution by giving particular values to the arbitrary constants. It is entirely free from arbitrary constants.

Formation of Differential Equations When a family of curves is defined by an equation containing $n$ independent arbitrary constants, the corresponding differential equation is obtained by differentiating the equation $n$ times and eliminating all arbitrary constants. The resulting differential equation will have an order equal to $n$. → [JEE TIP] Before differentiating, always simplify the equation to find the effective number of independent arbitrary constants (e.g., $A e^{x+B}$ simplifies to $C e^x$, so the order is 1, not 2).

Homogeneous Function A function $F(x, y)$ is said to be a homogeneous function of degree $n$ if substituting $\lambda x$ for $x$ and $\lambda y$ for $y$ yields $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ for any non-zero constant $\lambda$.

Geometrical Applications For any curve $y = f(x)$ at a point $(x,y)$, the derivative $dy/dx$ represents the slope of the tangent. Important terms frequently tested in JEE:

• Slope of Normal: $-dx/dy$
• Length of Tangent: $|y \sqrt{1 + (dx/dy)^2}|$
• Length of Normal: $|y \sqrt{1 + (dy/dx)^2}|$
• Length of Subtangent: $|y (dx/dy)|$
• Length of Subnormal: $|y (dy/dx)|$

Methods of Solving First Order First Degree Differential Equations

Variable Separable Method If the differential equation can be expressed as $dy/dx = g(x)h(y)$, the variables can be separated completely. The equation is rewritten as $\frac{1}{h(y)} dy = g(x) dx$ (provided $h(y) \neq 0$). Integrating both sides yields the general solution.

Equations Reducible to Variable Separable Differential equations of the form $dy/dx = f(ax + by + c)$ can be solved by making the substitution $ax + by + c = t$. Differentiating with respect to $x$ gives $a + b(dy/dx) = dt/dx$, which converts the equation into a variable separable form in $t$ and $x$. → [JEE TIP] This is a highly probable pattern in JEE Advanced matching type questions.

Homogeneous Differential Equations A differential equation of the form $dy/dx = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree zero.

• To solve $dy/dx = F(x, y) = g(y/x)$, make the substitution $y = vx$. Then $dy/dx = v + x(dv/dx)$.
• If the equation is of the form $dx/dy = F(x, y) = h(x/y)$, use the substitution $x = vy$. Then $dx/dy = v + y(dv/dy)$.

Equations Reducible to Homogeneous Form Equations of the form $\frac{dy}{dx} = \frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2}$ can be reduced:

• Case 1: If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, substitute $x = X + h$ and $y = Y + k$, choosing $h, k$ to eliminate the constant terms $c_1, c_2$. The equation becomes homogeneous in $X$ and $Y$.
• Case 2: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = m$, the equation contains a common linear factor. Substitute the repeating linear expression $a_2 x + b_2 y = t$ to reduce it to variable separable form.

Linear Differential Equations (LDE) A differential equation is linear if the dependent variable and its derivative appear only in the first degree and are not multiplied together. The standard form is: $\frac{dy}{dx} + Py = Q$ where $P$ and $Q$ are constants or functions of $x$ only. To solve, multiply both sides by the Integrating Factor (I.F.) $= e^{\int P dx}$. The solution is: $y \times (\text{I.F.}) = \int Q \times (\text{I.F.}) dx + C$.

Alternatively, the LDE can be in the form: $\frac{dx}{dy} + P_1 x = Q_1$ where $P_1$ and $Q_1$ are constants or functions of $y$ only. The I.F. $= e^{\int P_1 dy}$, and the solution is: $x \times (\text{I.F.}) = \int Q_1 \times (\text{I.F.}) dy + C$. → [JEE TIP] If finding $dy/dx$ leads to complex terms in the denominator, invert it to $dx/dy$ and check for this alternative linear form.

Equations Reducible to Linear Form (Bernoulli’s Equation) An equation of the form $\frac{dy}{dx} + Py = Qy^n$ is reducible to LDE. Divide the entire equation by $y^n$ to get $y^{-n} \frac{dy}{dx} + P y^{1-n} = Q$. Substitute $y^{1-n} = t$, which gives $(1-n) y^{-n} \frac{dy}{dx} = \frac{dt}{dx}$. The equation becomes linear in $t$ and $x$.

Exact Differential Equations An equation $M(x,y) dx + N(x,y) dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The solution is obtained by: $\int (\text{terms of } M \text{ treating } y \text{ as constant}) dx + \int (\text{terms of } N \text{ free from } x) dy = C$. → [JEE TIP] Many exact equations can be solved rapidly by grouping terms into exact differentials:

• $x dy + y dx = d(xy)$
• $\frac{x dy - y dx}{x^2} = d(\frac{y}{x})$
• $\frac{y dx - x dy}{y^2} = d(\frac{x}{y})$
• $\frac{x dy - y dx}{x^2 + y^2} = d(\tan^{-1}\frac{y}{x})$
• $\frac{x dx + y dy}{x^2 + y^2} = d(\frac{1}{2} \ln(x^2 + y^2))$

Orthogonal Trajectories A curve that cuts every member of a given family of curves at right angles. To find the orthogonal trajectory:

1. Form the differential equation of the given family: $f(x, y, y') = 0$.
2. Replace $y'$ ($dy/dx$) with $-dx/dy$.
3. Solve the new differential equation.

Conditions & Limitations

• Polynomial Restriction for Degree: The degree of a differential equation is strictly undefined if it cannot be algebraically manipulated into a polynomial of its derivatives. Example: $\cos(y') = x$ can be written as $y' = \cos^{-1}(x)$, so degree is 1. But $y'' + \cos(y') = 0$ cannot be isolated, so its degree is undefined.
• Variable Separation Failure: If an equation involves terms like $\sin(x+y)$, algebraic separation $g(x)h(y)$ is impossible, and substitution is mandatory.
• Zero Division in Separation: When dividing by $h(y)$ during variable separation (e.g., $dy/dx = -x/y$), we must assume $h(y) \neq 0$ for the derivation steps. The case where $h(y) = 0$ must be checked separately to see if it yields singular solutions.
• Function Constraints in LDE: In $\frac{dy}{dx} + Py = Q$, $P$ and $Q$ must strictly be constants or functions of $x$ only. The presence of $y$ inside $P$ or $Q$ makes it non-linear.
• Homogeneous Domain Limits: The substitution $y = vx$ is valid only if the function $F(x,y)$ truly factors out $\lambda^0$ under the transformation $(\lambda x, \lambda y)$.

Standard Derivations & Step-by-Step Problem Solving

Derivation of the Integrating Factor (I.F.) for LDE For a first-order linear differential equation: $\frac{dy}{dx} + P y = Q$

1. We want to find a function $g(x)$ such that multiplying both sides by $g(x)$ makes the L.H.S. an exact derivative of $(y \cdot g(x))$.
2. $g(x)\frac{dy}{dx} + P \cdot g(x)y = Q \cdot g(x)$.
3. We require the L.H.S to equal $\frac{d}{dx}[y \cdot g(x)] = g(x)\frac{dy}{dx} + y \cdot g'(x)$.
4. Equating the terms, we get $P \cdot g(x) = g'(x)$.
5. Separating variables: $P dx = \frac{g'(x)}{g(x)} dx$.
6. Integrating both sides: $\int P dx = \ln|g(x)|$.
7. Exponentiating: $g(x) = e^{\int P dx}$, which is defined as the Integrating Factor.

Step-by-Step Resolution of a Homogeneous Equation Example model: $x^2 dy + y(x+y) dx = 0$

1. Rearrange to standard form: $\frac{dy}{dx} = \frac{-y(x+y)}{x^2} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2$.
2. Substitute $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.
3. Substitute into the equation: $v + x\frac{dv}{dx} = -v - v^2$.
4. Isolate the $v$ derivative: $x\frac{dv}{dx} = -2v - v^2$.
5. Separate variables: $\frac{dv}{v(v+2)} = -\frac{dx}{x}$.
6. Integrate using partial fractions on the left, and log properties on the right.
7. Resubstitute $v = y/x$ to get the final implicit general solution.

Previous Year JEE Topics

• Bernoulli's Equation & Transformations: Equations needing division by $y^2$ or $y^3$ followed by substitution $1/y = t$ are heavily featured in JEE Advanced.
• Exact Differentials using Inspection: The ability to spot $x dy + y dx$ and combine it into $d(xy)$ to bypass lengthy homogeneous or linear methods is a critical time-saver.
• Curves with given Geometric Properties: Word problems stating "The subnormal is proportional to the square of the abscissa" translating to $|y \cdot y'| \propto x^2$, leading to a differential equation.
• Initial Value Problems (IVPs) in Physics contexts: Newton's Law of Cooling and Radioactive Decay modeled mathematically.
• Orthogonal Trajectories: Frequently tested in matrix match types. Finding the orthogonal trajectory to families like $x^2 + y^2 = 2ax$ or $xy = c^2$.

Memory Aids & JEE Traps

Based on this chapter, here are the top 10 MCQ traps, calculation tricks, and common misconceptions that appear in JEE:

Trap 1 - Polynomial Requirement for Degree

• [JEE TIP] Trap 1 - The Polynomial Derivative Constraint:

  • Misconception: The degree of the differential equation $y'' + \sin(y') = 0$ is equal to $1$ because the highest power of the highest-order derivative ($y''$) is $1$.
  • Correct Understanding: The degree of a differential equation is strictly undefined unless the equation can be expressed as a rational polynomial expression in terms of its derivatives. Because the first derivative is trapped inside a transcendental sine function ($\sin(y')$), it cannot be written as a polynomial, making its degree undefined.

• [JEE TIP] Trap 2 - The Pseudo-Higher Order Illusion:

  • Misconception: The differential equation corresponding to a family of curves with $n$ written constants always possesses an order of exactly $n$. For instance, $y = A \sin(x + B) + C$ has 3 constants, so its order is 3.
  • Correct Understanding: The order of a differential equation matches the number of essential, independent arbitrary constants. In $y = A \sin(x + B) + C$, the constants cannot be mathematically collapsed into one another, so the order is indeed 3. However, given an expression like $y = A e^{x} \cdot e^B$, it simplifies algebraically to $y = C e^x$ (where $C = Ae^B$). Always simplify parameters completely first; this dummy expression has an order of only 1.

• [JEE TIP] Trap 3 - The Flipped Linear Differential Equation:

  • Misconception: Every Linear Differential Equation (LDE) must strictly conform to the standard layout $\frac{dy}{dx} + P(x)y = Q(x)$, and any equation where $y$ is non-linear (like $\frac{dy}{dx} = \frac{1}{x+y^2}$) requires complex variable substitutions.
  • Correct Understanding: If an equation looks hopelessly non-linear in terms of $\frac{dy}{dx}$, invert the derivatives to look at it as a function of $\frac{dx}{dy}$. Flipping the equation yields $\frac{dx}{dy} = x + y^2 \Rightarrow \frac{dx}{dy} - x = y^2$. This is a standard, highly manageable LDE in $x$ where the Integrating Factor is calculated with respect to $y$: $\text{I.F.} = e^{\int -1 dy} = e^{-y}$.

• [JEE TIP] Trap 4 - The Integrating Factor Constant Redundancy:

  • Misconception: While evaluating the Integrating Factor $\text{I.F.} = e^{\int P dx}$, an arbitrary constant of integration ($+C$) must be explicitly added to the exponent.
  • Correct Understanding: No constant of integration is added during the intermediate $\text{I.F.}$ calculation step. Adding a constant creates an extra multiplier ($e^C$), which ultimately cancels out from both sides of the equation anyway. The single, mandatory constant of integration $+C$ must only be introduced at the final integration step: $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.})\, dx + C$.

• [JEE TIP] Trap 5 - The Modulus Omission Breakdown:

  • Misconception: The indefinite integration of the reciprocal function is simply expressed as $\int \frac{1}{x}\, dx = \ln(x)$.
  • Correct Understanding: The integration must strictly account for negative domains using a modulus: $\int \frac{1}{x}\, dx = \ln|x|$. In JEE boundary-value problems where a curve passes through a coordinate with negative values (such as $(-2, 3)$), omitting the modulus creates mathematically invalid imaginary logs ($\ln(-2)$) and breaks the entire calculation chain.

• [JEE TIP] Trap 6 - The Constant Homogeneous Destroyer:

  • Misconception: A differential expression like $\frac{dy}{dx} = \frac{x^2 + y^2 + 1}{xy}$ is classified as a homogeneous equation because the leading terms on the top and bottom are all quadratic.
  • Correct Understanding: This equation is not homogeneous. The presence of the standalone constant "$+1$" violates the fundamental scaling property $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$. For an equation to be homogeneous, every single individual term across the numerators and denominators must possess the exact same combined algebraic degree in $x$ and $y$.

• [JEE TIP] Trap 7 - The Lost Trivial Solution:

  • Misconception: Separating variables for $\frac{dy}{dx} = y$ into $\frac{dy}{y} = dx \Rightarrow \ln|y| = x + C \Rightarrow y = A e^x$ yields a complete general solution where the coefficient constraint is strictly $A \neq 0$.
  • Correct Understanding: The algebraic act of dividing both sides by $y$ introduces an implicit domain assumption that $y \neq 0$. Consequently, the trivial solution $y = 0$ is lost during the variable separation process. Because $y = 0$ perfectly satisfies the initial differential equation ($\frac{d(0)}{dx} = 0$), it is a valid solution and is integrated into the general form by allowing $A$ to equal $0$.

• [JEE TIP] Trap 8 - The Non-Unity Leading Derivative Trap:

  • Misconception: For the differential equation $x^2 \frac{dy}{dx} + x y = x^3$, the Integrating Factor can be directly evaluated from the coefficient of $y$ as $e^{\int x\, dx} = e^{x^2/2}$.
  • Correct Understanding: Before calculating the Integrating Factor, the coefficient of the highest derivative term $\frac{dy}{dx}$ must strictly be normalized to $1$. Dividing the entire equation by $x^2$ yields the proper form: $\frac{dy}{dx} + \frac{1}{x} y = x$. This reveals the true $P(x) = \frac{1}{x}$, resulting in a correct Integrating Factor of $\text{I.F.} = e^{\int (1/x)\, dx} = e^{\ln x} = x$.

• [JEE TIP] Trap 9 - The Orthogonal Trajectory Sign Deficit:

  • Misconception: To determine the differential equation of an orthogonal trajectory, you simply replace the derivative term $y'$ with its reciprocal $\frac{1}{y'}$.
  • Correct Understanding: Orthogonal trajectories intersect the original family of curves at a perfect $90^\circ$ angle, meaning their slopes must satisfy the perpendicular condition $m_1 m_2 = -1$. Therefore, you must replace the derivative $\frac{dy}{dx}$ with its negative reciprocal: $-\frac{dx}{dy}$. Dropping the negative sign produces completely incorrect trajectories.

• [JEE TIP] Trap 10 - The Exact Differential Mix-Up:

  • Misconception: The common variable pairing $x\,dx + y\,dy$ can be compressed instantly into the exact differential $d(xy)$.
  • Correct Understanding: The differential of a product follows the product rule: $d(xy) = x\,dy + y\,dx$. Conversely, the expression $x\,dx + y\,dy$ integrates into a sum of squares and represents the exact differential $d\left(\frac{x^2 + y^2}{2}\right)$. Confusing these two combinations will completely derail the integration phase of an exact differential equation problem.