2026?27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Order, degree, general solutions and particular solutions.
- Solving by separation of variables and first-order, first-degree homogeneous equations.
- Linear differential equations in x or y, as listed in the syllabus.
Last 24-Hour Checklist
Basic Concepts
Order & Degree
Order: The order of the highest order derivative appearing in the equation.
Degree: The power of the highest order derivative when the equation is a polynomial in derivatives.
Degree: The power of the highest order derivative when the equation is a polynomial in derivatives.
Degree Defined
$(\frac{d^2y}{dx^2})^3 + 5\frac{dy}{dx} = 0$
Order = 2, Degree = 3
Degree NOT Defined
$\frac{dy}{dx} + \sin(\frac{dy}{dx}) = 0$
Not polynomial in derivatives.
General & Particular Solutions
General Solution: Contains arbitrary constants equal to the order of DE.
Particular Solution: Obtained by giving specific values to constants (using initial conditions).
Particular Solution: Obtained by giving specific values to constants (using initial conditions).
Methods of Solving DE
1. Variable Separable
If $\frac{dy}{dx} = g(x) \cdot h(y)$, separate as $\frac{1}{h(y)} dy = g(x) dx$ and integrate.
2. Homogeneous Differential Equations
Form: $\frac{dy}{dx} = f(x,y)$ where $f(λ x, λ y) = λ^0 f(x,y)$
Step-by-Step Method:
Step-by-Step Method:
- Substitute $y = vx$.
- Differentiate with respect to $x$: $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
- Substitute $y$ and $\frac{dy}{dx}$ back into the original equation. The equation will reduce to variable separable form in $v$ and $x$.
- Separate variables, integrate both sides, and finally replace $v = \frac{y}{x}$.
3. Linear Differential Equations
Standard Form 1: $\frac{dy}{dx} + Py = Q$ (where $P, Q$ are functions of $x$ or
constants)
Integrating Factor (I.F.): $I.F. = e^{\int P dx}$
Solution: $y \cdot (I.F.) = \int (Q \cdot I.F.) \, dx + C$
Standard Form 2: $\frac{dx}{dy} + P_1x = Q_1$ (where $P_1, Q_1$ are functions of $y$ or constants)
Integrating Factor (I.F.): $I.F. = e^{\int P_1 dy}$
Solution: $x \cdot (I.F.) = \int (Q_1 \cdot I.F.) \, dy + C$
Integrating Factor (I.F.): $I.F. = e^{\int P dx}$
Solution: $y \cdot (I.F.) = \int (Q \cdot I.F.) \, dx + C$
Standard Form 2: $\frac{dx}{dy} + P_1x = Q_1$ (where $P_1, Q_1$ are functions of $y$ or constants)
Integrating Factor (I.F.): $I.F. = e^{\int P_1 dy}$
Solution: $x \cdot (I.F.) = \int (Q_1 \cdot I.F.) \, dy + C$
⚡ High-Yield Differential Equations Rules & Traps (MCQs & 3/5 Marks):
- Order & Degree Criteria (Frequent Board MCQ):
• Order: The highest derivative present in the equation. It is ALWAYS defined and is a positive integer.
• Degree: The highest power (exponent) of the highest order derivative, PROVIDED the differential equation can be written as a polynomial in all its derivatives.
• The Degree Undefined Trap: If any derivative is enclosed inside a transcendental function, such as $\sin\left(\frac{dy}{dx}\right)$, $e^{\frac{dy}{dx}}$, or $\ln\left(\frac{d^2y}{dx^2}\right)$, the Degree is NOT DEFINED (Order remains well-defined!). - Arbitrary Constants Rule (A-R & MCQ Favorite):
• The general solution of a differential equation of order $n$ contains exactly $n$ arbitrary constants.
• The particular solution of a differential equation of ANY order contains exactly $0$ arbitrary constants. - Linear Differential Equations — Both Forms:
• Type 1: $\frac{dy}{dx} + Py = Q$ ($P, Q$ functions of $x$): $$\text{IF} = e^{\int P \, dx} \implies y \cdot (\text{IF}) = \int [Q \cdot (\text{IF})] \, dx + C$$ • Type 2: $\frac{dx}{dy} + P_1x = Q_1$ ($P_1, Q_1$ functions of $y$): $$\text{IF} = e^{\int P_1 \, dy} \implies x \cdot (\text{IF}) = \int [Q_1 \cdot (\text{IF})] \, dy + C$$ • Simplification Rule: $e^{\ln f(x)} = f(x)$ and $e^{-\ln x} = e^{\ln(x^{-1})} = \frac{1}{x}$. - Homogeneous Differential Equations:
• A function satisfies $f(\lambda x, \lambda y) = \lambda^0 f(x, y)$.
• For $\frac{dy}{dx} = f(x, y)$, substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$.
• For $\frac{dx}{dy} = g(x, y)$, substitute $x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}$. - 📌 Syllabus Clarification: Formation of differential equations from general solutions (NCERT Ex 9.3) has been deleted from the active CBSE syllabus.
Solved Examples (Board Marking Scheme)
Q1. Solve: $\frac{dy}{dx} = \frac{1+x}{2-y}$ where $y \neq 2$. (2 Marks)
Step 1: Separate variables 1 Mark
$(2-y)dy = (1+x)dx$.
Step 2: Integrate 1 Mark
$\int (2-y) dy = \int (1+x) dx$
$2y - \frac{y^2}{2} = x + \frac{x^2}{2} + C$.
Q2. Solve: $(x-y)\frac{dy}{dx} = x + 2y$. (4 Marks)
Step 1: Identify Homogeneous 1 Mark
$\frac{dy}{dx} = \frac{x+2y}{x-y}$. Put $y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$.
Step 2: Separate Variables 1.5 Marks
$v + x\frac{dv}{dx} = \frac{1+2v}{1-v} \Rightarrow x\frac{dv}{dx} = \frac{1+2v-v+v^2}{1-v} = \frac{v^2+v+1}{1-v}$.
$\frac{1-v}{v^2+v+1} dv = \frac{dx}{x}$.
Step 3: Integrate & Substitute 1.5 Marks
Integrate both sides and replace $v = y/x$.
Q3. Solve: $\frac{dy}{dx} - y = \cos x$. (4 Marks)
Step 1: Find I.F. 1 Mark
$P = -1, Q = \cos x$.
$I.F. = e^{\int -1 dx} = e^{-x}$.
Step 2: Solution Formula 1 Mark
$y \cdot e^{-x} = \int e^{-x} \cos x dx$.
Step 3: Integrate RHS 2 Marks
Let $I = \int e^{-x} \cos x dx$. Use integration by parts twice.
$I = \frac{e^{-x}}{2}(\sin x - \cos x)$.
Final: $y = \frac{1}{2}(\sin x - \cos x) + Ce^x$.
Previous Year Questions (PYQs)
2023 (1 Mark MCQ): The degree of the differential equation
$(\frac{d^2y}{dx^2})^2 + (\frac{dy}{dx})^3 = 5$ is:
(A) 1 (B) 2 (C) 3 (D) Not defined
Ans: (B). Highest order is 2, its power is 2.
(A) 1 (B) 2 (C) 3 (D) Not defined
Ans: (B). Highest order is 2, its power is 2.
2022 (4 Marks): Solve the differential equation: $(x+y)\frac{dy}{dx} = 1$.
Hint: Rewrite as $\frac{dx}{dy} = x+y \Rightarrow \frac{dx}{dy} - x = y$. This is Linear DE in $x$.
$P = -1, Q = y$. $I.F. = e^{-y}$. Solution: $x e^{-y} = \int y e^{-y} dy$.
Hint: Rewrite as $\frac{dx}{dy} = x+y \Rightarrow \frac{dx}{dy} - x = y$. This is Linear DE in $x$.
$P = -1, Q = y$. $I.F. = e^{-y}$. Solution: $x e^{-y} = \int y e^{-y} dy$.
2019 (1 Mark MCQ): The integrating factor of $\frac{dy}{dx} + y\tan x = \sec x$
is:
(A) $\cos x$ (B) $\sec x$ (C) $e^{\tan x}$ (D) $\tan x$
Ans: (B). $I.F. = e^{\int \tan x dx} = e^{\log|\sec x|} = \sec x$.
(A) $\cos x$ (B) $\sec x$ (C) $e^{\tan x}$ (D) $\tan x$
Ans: (B). $I.F. = e^{\int \tan x dx} = e^{\log|\sec x|} = \sec x$.
Exam Strategy & Mistake Bank
Common Mistakes
Mistake 1: Confusing order and degree. Order = highest derivative, Degree =
power of highest derivative.
Mistake 2: Forgetting to add constant $C$ after integration.
Mistake 3: In homogeneous DE, forgetting to replace $v$ by $y/x$ in the
final answer.
Scoring Tips
Tip 1: Always write the type of DE before solving (Variable
Separable/Homogeneous/Linear).
Tip 2: For Linear DE, clearly write $P$, $Q$, and $I.F.$ - this shows your
method and earns partial marks.
Tip 3: For particular solution, substitute initial conditions and show
calculation of $C$.
Practice Problems (Self-Assessment)
Level 1: Basic (2 Marks Each)
Q1. Find order and degree: $(\frac{dy}{dx})^3 + 2\frac{dy}{dx} = x$.
Answer: Order = 1, Degree = 3.
Q2. Solve: $\frac{dy}{dx} = e^{x+y}$.
Answer: $e-y dy = e^x dx \Rightarrow -e-y = e^x + C \Rightarrow e^x +
e-y = K$.
Level 2: Intermediate (4 Marks Each)
Q3. Solve: $\frac{dy}{dx} + \frac{y}{x} = x^2$.
Hint: Linear DE. $P = 1/x, Q = x^2$. $I.F. = x$.
Solution: $yx = \int x^3 dx = x^4/4 + C$.
Solution: $yx = \int x^3 dx = x^4/4 + C$.
Q4. Solve: $(x^2+xy)dy = (x^2+y^2)dx$.
Hint: Homogeneous DE. Put $y=vx$.
Chapter 9 Board Mastery Test
Practice test • 10 marks • 25 min
Q1 (1 Mark): Find the order and degree (if defined) of the differential equation: $(\frac{d^2y}{dx^2})^2 + \cos(\frac{dy}{dx}) = 0$.
Q2 (2 Marks): Find the integrating factor of the differential equation $x\frac{dy}{dx} - y = 2x^2$.
Q3 (3 Marks): Find the particular solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$, given that $y = 0$ when $x = 0$.
Q4 (4 Marks): Solve the differential equation: $x\frac{dy}{dx} + 2y = x^2 \log x$ ($x > 0$).