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Differential Equations
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Class 12MathsChapter 9NCERTCBSE
Differential Equations - Class 12 Maths Chapter 9
📌 Quick Overview: Chapter 9 introduces differential equations — equations involving derivatives of a function. You will learn the order and degree of a differential equation, how to form a differential equation from a general solution, and three main methods to solve them: variable separable, homogeneous, and linear differential equations. This chapter is consistently tested in CBSE board exams — usually one 5-mark question and one 2-mark question every year.
Topics Covered:
- > What is a Differential Equation?
- > Order and Degree
- > General and Particular Solution
- > Formation of DE
- > Variable Separable Method
- > Homogeneous DE
- > Linear DE (Integrating Factor)
- > Bernoulli's Equation
- > Applications of DE
- > Important Board Questions
- > MCQ Practice
1. What is a Differential Equation?
Definition: An equation involving the independent variable (x), dependent variable (y), and derivatives of y with respect to x is called a differential equation (DE).
Examples:
> dy/dx = x + y
> d^2y/dx^2 + 4y = 0
> (dy/dx)^2 + dy/dx = sin x
Examples:
> dy/dx = x + y
> d^2y/dx^2 + 4y = 0
> (dy/dx)^2 + dy/dx = sin x
2. Order and Degree of a Differential Equation ⭐
Order: The order of a DE is the order of the highest derivative present in the equation.
Degree: The degree of a DE is the power (exponent) of the highest order derivative, after the equation is made free of radicals and fractions in derivatives.
Degree: The degree of a DE is the power (exponent) of the highest order derivative, after the equation is made free of radicals and fractions in derivatives.
⭐ Important: Degree is defined ONLY when the DE is a polynomial in derivatives. If the DE has sin(dy/dx), e^(d^2y/dx^2), log(dy/dx), etc., the degree is NOT defined.
| Differential Equation | Order | Degree |
|---|---|---|
| dy/dx = x^2 + y | 1 | 1 |
| (dy/dx)^3 + 4y = sin x | 1 | 3 |
| d^2y/dx^2 + 3(dy/dx) + y = 0 | 2 | 1 |
| (d^2y/dx^2)^2 + dy/dx = x | 2 | 2 |
| sqrt(1 + (dy/dx)^2) = d^2y/dx^2 | 2 | 2 (square both sides first) |
| d^2y/dx^2 + sin(dy/dx) = 0 | 2 | Not Defined |
| dy/dx + e^y = x | 1 | Not Defined |
3. General and Particular Solution
General Solution: The solution of a DE that contains arbitrary constants (as many as the order of the DE) is called the general solution.
Particular Solution: A solution obtained by giving specific values to the arbitrary constants in the general solution (using given initial conditions) is called a particular solution.
Particular Solution: A solution obtained by giving specific values to the arbitrary constants in the general solution (using given initial conditions) is called a particular solution.
Example: y = A sin x + B cos x is the general solution of d^2y/dx^2 + y = 0 (two constants A and B for order 2).
If y(0) = 1 and y'(0) = 0, then A = 0, B = 1 → particular solution: y = cos x.
If y(0) = 1 and y'(0) = 0, then A = 0, B = 1 → particular solution: y = cos x.
4. Formation of a Differential Equation ⭐
Method: To form a DE from a given general solution containing n arbitrary constants, differentiate n times and eliminate the n arbitrary constants.
Steps:
Step 1: Count the number of arbitrary constants (n).
Step 2: Differentiate n times.
Step 3: Eliminate the arbitrary constants using the equations obtained.
Step 4: Write the final DE in terms of x, y, and derivatives only.
Step 1: Count the number of arbitrary constants (n).
Step 2: Differentiate n times.
Step 3: Eliminate the arbitrary constants using the equations obtained.
Step 4: Write the final DE in terms of x, y, and derivatives only.
Example 1: Form the DE from y = Ae^x + Be^(-x).
Two constants A and B → differentiate twice.
y' = Ae^x - Be^(-x) ... (1)
y'' = Ae^x + Be^(-x) ... (2)
From (2): y'' = y (since Ae^x + Be^(-x) = y)
DE: d^2y/dx^2 - y = 0
Two constants A and B → differentiate twice.
y' = Ae^x - Be^(-x) ... (1)
y'' = Ae^x + Be^(-x) ... (2)
From (2): y'' = y (since Ae^x + Be^(-x) = y)
DE: d^2y/dx^2 - y = 0
Example 2: Form the DE from y = mx where m is an arbitrary constant.
One constant → differentiate once.
dy/dx = m ... (1)
From y = mx: m = y/x
Substitute in (1): dy/dx = y/x
DE: x(dy/dx) = y
One constant → differentiate once.
dy/dx = m ... (1)
From y = mx: m = y/x
Substitute in (1): dy/dx = y/x
DE: x(dy/dx) = y
Example 3: Form the DE from y = a sin(x + b) where a and b are constants.
Two constants → differentiate twice.
y' = a cos(x+b) ... (1)
y'' = -a sin(x+b) = -y ... (2)
DE: d^2y/dx^2 + y = 0
Two constants → differentiate twice.
y' = a cos(x+b) ... (1)
y'' = -a sin(x+b) = -y ... (2)
DE: d^2y/dx^2 + y = 0
5. Method 1 - Variable Separable ⭐
Concept: If a DE can be written in the form f(x)dx = g(y)dy, then all x terms go to one side and y terms to the other. Integrate both sides to get the solution.
Steps:
Step 1: Rearrange the DE so that all x terms (with dx) are on one side and all y terms (with dy) on the other.
Step 2: Integrate both sides.
Step 3: Add constant of integration C on one side.
Step 4: If initial condition given, find C for particular solution.
Step 1: Rearrange the DE so that all x terms (with dx) are on one side and all y terms (with dy) on the other.
Step 2: Integrate both sides.
Step 3: Add constant of integration C on one side.
Step 4: If initial condition given, find C for particular solution.
Example 1: Solve dy/dx = (1+y^2)/(1+x^2)
Separating: dy/(1+y^2) = dx/(1+x^2)
Integrating both sides:
tan⁻¹(y) = tan⁻¹(x) + C
tan⁻¹(y) - tan⁻¹(x) = C
Separating: dy/(1+y^2) = dx/(1+x^2)
Integrating both sides:
tan⁻¹(y) = tan⁻¹(x) + C
tan⁻¹(y) - tan⁻¹(x) = C
Example 2: Solve dy/dx = e^(x+y)
dy/dx = e^x . e^y
e^(-y) dy = e^x dx
Integrating: -e^(-y) = e^x + C
e^x + e^(-y) = -C (or K, a new constant)
dy/dx = e^x . e^y
e^(-y) dy = e^x dx
Integrating: -e^(-y) = e^x + C
e^x + e^(-y) = -C (or K, a new constant)
Example 3: Solve dy/dx = (x(2 log x + 1)) / (sin y + y cos y)
(sin y + y cos y) dy = x(2 log x + 1) dx
Left: Integral of (sin y + y cos y) dy = -cos y + y sin y (by parts on second term)
Right: Integral of x(2 log x + 1)dx = x^2 log x (verify by differentiation)
y sin y = x^2 log x + C
(sin y + y cos y) dy = x(2 log x + 1) dx
Left: Integral of (sin y + y cos y) dy = -cos y + y sin y (by parts on second term)
Right: Integral of x(2 log x + 1)dx = x^2 log x (verify by differentiation)
y sin y = x^2 log x + C
Example 4 (Particular Solution): Solve dy/dx = y tan x, given y = 1 when x = 0.
dy/y = tan x dx
Integrating: log|y| = log|sec x| + C
y = A sec x (where A = e^C)
At x = 0, y = 1: 1 = A.sec(0) = A.1 → A = 1
Particular Solution: y = sec x
dy/y = tan x dx
Integrating: log|y| = log|sec x| + C
y = A sec x (where A = e^C)
At x = 0, y = 1: 1 = A.sec(0) = A.1 → A = 1
Particular Solution: y = sec x
6. Method 2 - Homogeneous Differential Equations ⭐
Definition: A DE of the form dy/dx = F(x,y) is called homogeneous if F(x,y) can be written as a function of (y/x) alone, i.e., F(x,y) = g(y/x).
Equivalently: F(tx, ty) = F(x,y) for any t — all terms have the same total degree.
Equivalently: F(tx, ty) = F(x,y) for any t — all terms have the same total degree.
Method to Solve Homogeneous DE:
Step 1: Verify the DE is homogeneous.
Step 2: Substitute y = vx, so dy/dx = v + x(dv/dx).
Step 3: Substitute into DE — it becomes variable separable in v and x.
Step 4: Solve by separating variables.
Step 5: Substitute back v = y/x to get solution in x and y.
Step 1: Verify the DE is homogeneous.
Step 2: Substitute y = vx, so dy/dx = v + x(dv/dx).
Step 3: Substitute into DE — it becomes variable separable in v and x.
Step 4: Solve by separating variables.
Step 5: Substitute back v = y/x to get solution in x and y.
Example 1: Solve (x^2 + xy)dy = (x^2 + y^2)dx
dy/dx = (x^2 + y^2)/(x^2 + xy) = (1 + (y/x)^2)/(1 + y/x)
This is a function of y/x only → Homogeneous ✓
Let y = vx → dy/dx = v + x(dv/dx)
v + x(dv/dx) = (1 + v^2)/(1 + v)
x(dv/dx) = (1+v^2)/(1+v) - v = (1+v^2 - v - v^2)/(1+v) = (1-v)/(1+v)
Separating: (1+v)/(1-v) dv = dx/x
[2/(1-v) - 1] dv = dx/x
-2 log|1-v| - v = log|x| + C
Substituting v = y/x:
-2 log|1 - y/x| - y/x = log|x| + C
-2 log|(x-y)/x| - y/x = log|x| + C
-2log|x-y| + 2log|x| - y/x = log|x| + C
dy/dx = (x^2 + y^2)/(x^2 + xy) = (1 + (y/x)^2)/(1 + y/x)
This is a function of y/x only → Homogeneous ✓
Let y = vx → dy/dx = v + x(dv/dx)
v + x(dv/dx) = (1 + v^2)/(1 + v)
x(dv/dx) = (1+v^2)/(1+v) - v = (1+v^2 - v - v^2)/(1+v) = (1-v)/(1+v)
Separating: (1+v)/(1-v) dv = dx/x
[2/(1-v) - 1] dv = dx/x
-2 log|1-v| - v = log|x| + C
Substituting v = y/x:
-2 log|1 - y/x| - y/x = log|x| + C
-2 log|(x-y)/x| - y/x = log|x| + C
-2log|x-y| + 2log|x| - y/x = log|x| + C
Example 2: Solve x dy/dx - y = sqrt(x^2 + y^2)
dy/dx = [y + sqrt(x^2+y^2)] / x = y/x + sqrt(1 + (y/x)^2)
Homogeneous ✓
Let y = vx: v + x(dv/dx) = v + sqrt(1+v^2)
x(dv/dx) = sqrt(1+v^2)
dv/sqrt(1+v^2) = dx/x
Integrating: log|v + sqrt(1+v^2)| = log|x| + C
v + sqrt(1+v^2) = Ax
Substituting v = y/x:
y/x + sqrt(1 + y^2/x^2) = Ax
y + sqrt(x^2+y^2) = Ax^2
dy/dx = [y + sqrt(x^2+y^2)] / x = y/x + sqrt(1 + (y/x)^2)
Homogeneous ✓
Let y = vx: v + x(dv/dx) = v + sqrt(1+v^2)
x(dv/dx) = sqrt(1+v^2)
dv/sqrt(1+v^2) = dx/x
Integrating: log|v + sqrt(1+v^2)| = log|x| + C
v + sqrt(1+v^2) = Ax
Substituting v = y/x:
y/x + sqrt(1 + y^2/x^2) = Ax
y + sqrt(x^2+y^2) = Ax^2
7. Method 3 - Linear Differential Equations ⭐ (Most Important)
Standard Form: dy/dx + P(x).y = Q(x)
where P and Q are functions of x only (or constants).
Integrating Factor (IF): IF = e^(Integral of P dx)
Solution: y . IF = Integral of (Q . IF) dx + C
where P and Q are functions of x only (or constants).
Integrating Factor (IF): IF = e^(Integral of P dx)
Solution: y . IF = Integral of (Q . IF) dx + C
Steps for Linear DE:
Step 1: Write DE in standard form: dy/dx + Py = Q.
Step 2: Identify P and Q (functions of x).
Step 3: Find IF = e^(Integral P dx).
Step 4: Multiply both sides by IF.
Step 5: Left side becomes d/dx(y.IF).
Step 6: Integrate both sides: y.IF = Integral(Q.IF)dx + C.
Step 7: Solve for y.
Step 1: Write DE in standard form: dy/dx + Py = Q.
Step 2: Identify P and Q (functions of x).
Step 3: Find IF = e^(Integral P dx).
Step 4: Multiply both sides by IF.
Step 5: Left side becomes d/dx(y.IF).
Step 6: Integrate both sides: y.IF = Integral(Q.IF)dx + C.
Step 7: Solve for y.
⭐ Also valid for: dx/dy + P(y).x = Q(y)
IF = e^(Integral P dy)
Solution: x.IF = Integral(Q.IF)dy + C
When to use: if the equation is easier to write as x as a function of y.
IF = e^(Integral P dy)
Solution: x.IF = Integral(Q.IF)dy + C
When to use: if the equation is easier to write as x as a function of y.
Example 1: Solve dy/dx + y/x = x^2
Standard form: dy/dx + (1/x)y = x^2
P = 1/x, Q = x^2
IF = e^(Integral 1/x dx) = e^(log x) = x
Multiply by IF: x(dy/dx) + y = x^3
d/dx(xy) = x^3
Integrating: xy = x^4/4 + C
y = x^3/4 + C/x
Standard form: dy/dx + (1/x)y = x^2
P = 1/x, Q = x^2
IF = e^(Integral 1/x dx) = e^(log x) = x
Multiply by IF: x(dy/dx) + y = x^3
d/dx(xy) = x^3
Integrating: xy = x^4/4 + C
y = x^3/4 + C/x
Example 2: Solve dy/dx + 2y tan x = sin x
P = 2 tan x, Q = sin x
IF = e^(Integral 2 tan x dx) = e^(2 log|sec x|) = sec^2 x
Multiply by sec^2 x:
d/dx(y sec^2 x) = sin x . sec^2 x = sin x/cos^2 x = tan x . sec x
Integrating: y sec^2 x = sec x + C
y = cos x + C cos^2 x
P = 2 tan x, Q = sin x
IF = e^(Integral 2 tan x dx) = e^(2 log|sec x|) = sec^2 x
Multiply by sec^2 x:
d/dx(y sec^2 x) = sin x . sec^2 x = sin x/cos^2 x = tan x . sec x
Integrating: y sec^2 x = sec x + C
y = cos x + C cos^2 x
Example 3 (x as function of y): Solve (1+y^2)dx = (tan⁻¹y - x)dy
Rearrange: dx/dy = (tan⁻¹y - x)/(1+y^2)
dx/dy + x/(1+y^2) = tan⁻¹y/(1+y^2)
P = 1/(1+y^2), Q = tan⁻¹y/(1+y^2)
IF = e^(Integral 1/(1+y^2) dy) = e^(tan⁻¹y)
x . e^(tan⁻¹y) = Integral [tan⁻¹y/(1+y^2)] . e^(tan⁻¹y) dy
Let t = tan⁻¹y, dt = dy/(1+y^2):
= Integral t.e^t dt = te^t - e^t + C = e^t(t-1) + C
x.e^(tan⁻¹y) = e^(tan⁻¹y)(tan⁻¹y - 1) + C
x = (tan⁻¹y - 1) + C.e^(-tan⁻¹y)
Rearrange: dx/dy = (tan⁻¹y - x)/(1+y^2)
dx/dy + x/(1+y^2) = tan⁻¹y/(1+y^2)
P = 1/(1+y^2), Q = tan⁻¹y/(1+y^2)
IF = e^(Integral 1/(1+y^2) dy) = e^(tan⁻¹y)
x . e^(tan⁻¹y) = Integral [tan⁻¹y/(1+y^2)] . e^(tan⁻¹y) dy
Let t = tan⁻¹y, dt = dy/(1+y^2):
= Integral t.e^t dt = te^t - e^t + C = e^t(t-1) + C
x.e^(tan⁻¹y) = e^(tan⁻¹y)(tan⁻¹y - 1) + C
x = (tan⁻¹y - 1) + C.e^(-tan⁻¹y)
Example 4 (Particular Solution): Solve dy/dx - 3y = sin 2x, given y(0) = 0.
P = -3, Q = sin 2x
IF = e^(Integral -3 dx) = e^(-3x)
y.e^(-3x) = Integral sin(2x).e^(-3x) dx
Using IBP (or formula for Integral e^(ax) sin(bx)dx = e^(ax)(a sin bx - b cos bx)/(a^2+b^2)):
= e^(-3x)(-3 sin 2x - 2 cos 2x)/13 + C
y.e^(-3x) = e^(-3x)(-3 sin 2x - 2 cos 2x)/13 + C
y = (-3 sin 2x - 2 cos 2x)/13 + Ce^(3x)
At x=0, y=0: 0 = (0-2)/13 + C → C = 2/13
y = (-3 sin 2x - 2 cos 2x + 2e^(3x))/13
P = -3, Q = sin 2x
IF = e^(Integral -3 dx) = e^(-3x)
y.e^(-3x) = Integral sin(2x).e^(-3x) dx
Using IBP (or formula for Integral e^(ax) sin(bx)dx = e^(ax)(a sin bx - b cos bx)/(a^2+b^2)):
= e^(-3x)(-3 sin 2x - 2 cos 2x)/13 + C
y.e^(-3x) = e^(-3x)(-3 sin 2x - 2 cos 2x)/13 + C
y = (-3 sin 2x - 2 cos 2x)/13 + Ce^(3x)
At x=0, y=0: 0 = (0-2)/13 + C → C = 2/13
y = (-3 sin 2x - 2 cos 2x + 2e^(3x))/13
8. Applications of Differential Equations
8.1 Growth and Decay
If a quantity N grows or decays at a rate proportional to itself:
dN/dt = kN
Solution: N = N₀ e^(kt)
k > 0 → Growth; k < 0 → Decay
dN/dt = kN
Solution: N = N₀ e^(kt)
k > 0 → Growth; k < 0 → Decay
Example: Bacteria doubles every 3 hours. Find the time to become 8 times.
dN/dt = kN → N = N₀ e^(kt)
At t=3: 2N₀ = N₀ e^(3k) → e^(3k) = 2 → 3k = log 2 → k = log2/3
For N = 8N₀: 8N₀ = N₀ e^(kt) → e^(kt) = 8 = 2^3
kt = 3 log 2 → t = 3 log 2 / k = 3 log 2 / (log 2/3) = 9 hours
Answer: 9 hours
dN/dt = kN → N = N₀ e^(kt)
At t=3: 2N₀ = N₀ e^(3k) → e^(3k) = 2 → 3k = log 2 → k = log2/3
For N = 8N₀: 8N₀ = N₀ e^(kt) → e^(kt) = 8 = 2^3
kt = 3 log 2 → t = 3 log 2 / k = 3 log 2 / (log 2/3) = 9 hours
Answer: 9 hours
8.2 Newton's Law of Cooling
The rate of cooling of a body is proportional to the difference between its temperature and the surrounding temperature:
dT/dt = -k(T - T₀)
Solution: T - T₀ = (T_initial - T₀) e^(-kt)
dT/dt = -k(T - T₀)
Solution: T - T₀ = (T_initial - T₀) e^(-kt)
8.3 Geometrical Applications
For a curve y = f(x):
> Slope of tangent = dy/dx
> Equation of tangent at (x,y): Y - y = (dy/dx)(X - x)
> x-intercept of tangent = x - y(dx/dy)
> y-intercept of tangent = y - x(dy/dx)
> Slope of tangent = dy/dx
> Equation of tangent at (x,y): Y - y = (dy/dx)(X - x)
> x-intercept of tangent = x - y(dx/dy)
> y-intercept of tangent = y - x(dy/dx)
Example: Find the curve passing through (1,0) such that the slope of tangent at any point (x,y) is (y-1)/(x+1).
dy/dx = (y-1)/(x+1)
Variable separable: dy/(y-1) = dx/(x+1)
log|y-1| = log|x+1| + C
y-1 = A(x+1)
At (1,0): -1 = 2A → A = -1/2
y - 1 = -(x+1)/2 → 2y - 2 = -x - 1 → x + 2y = 1
dy/dx = (y-1)/(x+1)
Variable separable: dy/(y-1) = dx/(x+1)
log|y-1| = log|x+1| + C
y-1 = A(x+1)
At (1,0): -1 = 2A → A = -1/2
y - 1 = -(x+1)/2 → 2y - 2 = -x - 1 → x + 2y = 1
9. Important Board Exam Questions
Q1. Find the order and degree of: [1 + (dy/dx)^2]^(3/2) = d^2y/dx^2
Square both sides to remove radical:
[1 + (dy/dx)^2]^3 = (d^2y/dx^2)^2
Highest derivative: d^2y/dx^2 → Order = 2
Power of d^2y/dx^2: 2 → Degree = 2
[1 + (dy/dx)^2]^3 = (d^2y/dx^2)^2
Highest derivative: d^2y/dx^2 → Order = 2
Power of d^2y/dx^2: 2 → Degree = 2
Q2. Form the DE of family of circles touching the x-axis at origin.
Equation of family: x^2 + (y-a)^2 = a^2 (centre at (0,a))
x^2 + y^2 - 2ay = 0 → a = (x^2+y^2)/(2y)
Differentiating x^2 + y^2 - 2ay = 0:
2x + 2y(dy/dx) - 2a(dy/dx) = 0
Substitute a: 2x + 2y(dy/dx) - (x^2+y^2)/y . (dy/dx) = 0
Multiply by y: 2xy + (2y^2 - x^2 - y^2)(dy/dx) = 0
DE: 2xy dx + (y^2 - x^2) dy = 0
x^2 + y^2 - 2ay = 0 → a = (x^2+y^2)/(2y)
Differentiating x^2 + y^2 - 2ay = 0:
2x + 2y(dy/dx) - 2a(dy/dx) = 0
Substitute a: 2x + 2y(dy/dx) - (x^2+y^2)/y . (dy/dx) = 0
Multiply by y: 2xy + (2y^2 - x^2 - y^2)(dy/dx) = 0
DE: 2xy dx + (y^2 - x^2) dy = 0
Q3. Solve: (x + y)(dx - dy) = dx + dy
(x+y)dx - (x+y)dy = dx + dy
(x+y-1)dx = (x+y+1)dy
dy/dx = (x+y-1)/(x+y+1)
Let v = x+y, dv/dx = 1 + dy/dx
dy/dx = dv/dx - 1
dv/dx - 1 = (v-1)/(v+1)
dv/dx = (v-1)/(v+1) + 1 = 2v/(v+1)
(v+1)/(2v) dv = dx
(1/2 + 1/(2v)) dv = dx
Integrating: v/2 + (1/2)log|v| = x + C
(x+y)/2 + (1/2)log|x+y| = x + C
(y-x)/2 + (1/2)log|x+y| = C
(x+y-1)dx = (x+y+1)dy
dy/dx = (x+y-1)/(x+y+1)
Let v = x+y, dv/dx = 1 + dy/dx
dy/dx = dv/dx - 1
dv/dx - 1 = (v-1)/(v+1)
dv/dx = (v-1)/(v+1) + 1 = 2v/(v+1)
(v+1)/(2v) dv = dx
(1/2 + 1/(2v)) dv = dx
Integrating: v/2 + (1/2)log|v| = x + C
(x+y)/2 + (1/2)log|x+y| = x + C
(y-x)/2 + (1/2)log|x+y| = C
Q4. Solve the linear DE: (x^2 - 1)(dy/dx) + 2xy = 1/(x^2-1)
Divide by (x^2-1): dy/dx + 2x/(x^2-1) . y = 1/(x^2-1)^2
P = 2x/(x^2-1), Q = 1/(x^2-1)^2
IF = e^(Integral 2x/(x^2-1) dx) = e^(log|x^2-1|) = (x^2-1)
y(x^2-1) = Integral [1/(x^2-1)^2 . (x^2-1)] dx = Integral 1/(x^2-1) dx
= (1/2) log|(x-1)/(x+1)| + C
y(x^2-1) = (1/2) log|(x-1)/(x+1)| + C
P = 2x/(x^2-1), Q = 1/(x^2-1)^2
IF = e^(Integral 2x/(x^2-1) dx) = e^(log|x^2-1|) = (x^2-1)
y(x^2-1) = Integral [1/(x^2-1)^2 . (x^2-1)] dx = Integral 1/(x^2-1) dx
= (1/2) log|(x-1)/(x+1)| + C
y(x^2-1) = (1/2) log|(x-1)/(x+1)| + C
Q5. Find the curve for which the sum of x and y intercepts of the tangent equals the length of the normal.
x-intercept of tangent = x - y(dx/dy)
y-intercept of tangent = y - x(dy/dx)
Sum = x + y - y(dx/dy) - x(dy/dx)
Length of normal = y . sqrt(1 + (dy/dx)^2)
This leads to a differential equation solved by substitution. The curve is a circle: x^2 + y^2 = c^2.
y-intercept of tangent = y - x(dy/dx)
Sum = x + y - y(dx/dy) - x(dy/dx)
Length of normal = y . sqrt(1 + (dy/dx)^2)
This leads to a differential equation solved by substitution. The curve is a circle: x^2 + y^2 = c^2.
10. Key Formulas at a Glance
ORDER: highest derivative present
DEGREE: power of highest derivative (after clearing radicals)
FORMATION: differentiate n times (n = no. of constants), eliminate constants
METHOD 1 - VARIABLE SEPARABLE:
Rearrange to f(y)dy = g(x)dx, then integrate both sides.
METHOD 2 - HOMOGENEOUS:
Check: dy/dx = F(y/x)
Substitute y = vx → dy/dx = v + x(dv/dx)
Separate variables in v and x, integrate, substitute back v = y/x
METHOD 3 - LINEAR DE:
Standard form: dy/dx + Py = Q
IF = e^(Integral P dx)
Solution: y.IF = Integral(Q.IF)dx + C
OR: dx/dy + Px = Q (x as function of y)
IF = e^(Integral P dy)
Solution: x.IF = Integral(Q.IF)dy + C
GROWTH/DECAY: dN/dt = kN → N = N₀ e^(kt)
11. MCQ Practice (1 Mark)
1. The order and degree of (d^2y/dx^2)^3 + (dy/dx)^2 + sin(dy/dx) + 1 = 0 are:
(a) order 2, degree 3 (b) order 2, degree not defined (c) order 1, degree 2 (d) order 2, degree 2
Answer: (b) order 2, degree not defined — because of sin(dy/dx)
2. The general solution of dy/dx = e^(x+y) is:
(a) e^x + e^y = C (b) e^x - e^(-y) = C (c) e^(-y) = -e^x + C (d) e^x + e^(-y) = C
Answer: (d) e^x + e^(-y) = C
3. Integrating factor of dy/dx + y/x = x^2 is:
(a) e^x (b) log x (c) x (d) 1/x
Answer: (c) x — IF = e^(Integral 1/x dx) = e^(log x) = x
4. The number of arbitrary constants in the general solution of a DE of order 3 is:
(a) 1 (b) 2 (c) 3 (d) 0
Answer: (c) 3
5. Which of the following is a homogeneous DE?
(a) (x^2+y)dx = (y-x)dy (b) (x^2+y^2)dx = 2xy dy (c) y dx + (x+y^2)dy = 0 (d) (x+2y^3)dy = y dx
Answer: (b) (x^2+y^2)dx = 2xy dy — dy/dx = (x^2+y^2)/2xy is a function of y/x
12. Exam Tips
- For order and degree: always simplify (remove radicals, fractions in derivatives) BEFORE determining the degree.
- Degree is NOT defined if the DE has trig functions, exponentials, or logarithms of derivatives — state clearly "degree not defined."
- For formation problems: count constants first — that tells you how many times to differentiate.
- For homogeneous DE: after substituting y = vx, the equation MUST become variable separable — if it doesn't, recheck your substitution.
- For linear DE: always bring the equation to standard form dy/dx + Py = Q FIRST before identifying P and Q.
- Common mistake in IF: IF = e^(Integral P dx), NOT e^P dx. Integrate P first, then take e to the power.
- After finding general solution, always substitute the initial condition carefully to find C for particular solution questions.
- For application problems: set up the DE first (translate the word problem into an equation), then solve — don't jump to integration directly.
✅ Summary: Differential Equations has three core solving methods — master all three. Variable Separable is the easiest and most common. Homogeneous always uses y = vx substitution. Linear always uses the Integrating Factor formula. In CBSE board exams, one 5-mark question from this chapter is almost certain — it will be either a linear DE with a particular solution, or a homogeneous DE. Practice one full problem of each type every day.
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