JEE Main · Mathematics ↓ Falling

Differential Equations appeared 46 times across 3 years — 5.3% of Mathematics. This question is from Linear Differential Equations.

Year 2026 2025 2024 Total
Questions 13 17 16 46

Let y = y(x) be the solution of the differential equation (dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³. Then y((π)/(4)) is equal to

Solution & Explanation

Related Formula

For a first-order linear differential equation (dy)/(dx) + P(x)y = Q(x):

  • Integrating Factor (I.F.) = e∫ P(x) dx
  • Solution is y · I.F. = ∫ Q(x) · I.F. dx + C
Core Logic

Let's simplify the coefficient of y:

3 ² x + 3 = 3( ² x + 1) = 3 ² x

Thus, the equation is:

(dy)/(dx) + 3( ² x)y = ² x
Step 1: Finding Integrating Factor and General Solution

I.F. = e∫ 3 ² x dx = e3 x

The general solution is:

y · e3 x = ∫ ² x · e3 x dx + C

Substitute u = 3 x du = 3 ² x dx:

y · e3 x = (1)/(3) ∫ e^u du + C = (1)/(3) e3 x + C
Step 2: Solving for boundary conditions

Given y(0) = (1)/(3) + e³:

((1)/(3) + e³) · e⁰ = (1)/(3) e⁰ + C C = e³

Thus, the explicit function is:

y = (1)/(3) + e3 - 3 x

Evaluating at x = (π)/(4):

y((π)/(4)) = (1)/(3) + e3 - 3 (π/4) = (1)/(3) + e³⁻³ = (1)/(3) + 1 = (4)/(3)
Pattern Recognition

Recognizing that 3 ² x + 3 = 3 ² x converts the system immediately into a classic linear differential equation where the coefficient of y is exactly the derivative of the exponent of I.F. This makes integration virtually instantaneous.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

More Differential Equations Previous-Year Questions — Page 3

Q16 jee_main_2026_24_january_evening Linear Differential Equations
Let y = y(x) be a differentiable function in the interval (0, ∞) such that y(1) = 2 and t → x ( (t² y(x) - x² y(t))/(x - t) ) = 3 for each x > 0. Then 2y(2) is equal to
  • A. 18
  • B. 23
  • C. 27
  • D. 12

Solution

Related Formula
L'Hopital's Rule for (0)/(0) form: t → x (f(t))/(g(t)) = t → x (f'(t))/(g'(t)) Linear DE standard form: (dy)/(dx) + P(x)y = Q(x)
Core Logic

Evaluate the limit using L'Hopital's rule, differentiating with respect to t (treating x as a constant).

t → x ((d)/(dt) (t² y(x) - x² y(t)))/((d)/(dt)(x - t)) = 3 t → x (2t · y(x) - x² y'(t))/(-1) = 3

Substitute t = x:

(2x · y(x) - x² y'(x))/(-1) = 3 x² y'(x) - 2x y(x) = 3

Rearrange into standard linear differential equation format:

(dy)/(dx) - (2)/(x) y = (3)/(x²)
Step 1: Integrating Factor
I.F. = e∫ P dx = e∫ -(2)/(x) dx = e-2ln x = x⁻² = (1)/(x²)
Step 2: Solving the Differential Equation

Multiply the entire equation by I.F.:

y · ((1)/(x²)) = ∫ (3)/(x²) · (1)/(x²) dx (y)/(x²) = ∫ 3x⁻⁴ dx = 3 ( x⁻³-3) + c (y)/(x²) = -(1)/(x³) + c y(x) = cx² - (1)/(x)
Step 3: Using the Boundary Condition

Given y(1) = 2:

2 = c(1)² - (1)/(1) 2 = c - 1 c = 3

Thus, the function is y(x) = 3x² - (1)/(x).

Step 4: Evaluating the Target Value

Find y(2):

y(2) = 3(2)² - (1)/(2) = 12 - (1)/(2) = (23)/(2)

The question asks for 2y(2):

2y(2) = 2 ((23)/(2)) = 23
Pattern Recognition

A limit expression resembling Newton's difference quotient applied to a function block almost universally unwraps into a first-order linear differential equation via L'Hopital's rule with respect to the limit dummy variable.

Chapter Mix

Class 12 Maths: Differential Equations Class 11 Maths: Limits and Derivatives

Q20 jee_main_2026_28_january_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation x (dy)/(dx) - 2y = x³(2 - x³) ² y, x ≠ 0. If y(2) = 0, then (y(1)) is equal to
  • A. (3)/(4)
  • B. (7)/(4)
  • C. -(7)/(4)
  • D. -(3)/(4)

Solution

Core Logic

Linear Differential Equations
Linear Differential Equations
Given differential equation:

x (dy)/(dx) - 2y = x³(2 - x³) ² y

Divide both sides by ² y:

x ² y (dy)/(dx) - (2 y y)/( ² y) = x³(2 - x³) x ² y (dy)/(dx) - 2 y = x³(2 - x³)

Divide by x:

² y (dy)/(dx) - (2)/(x) y = x²(2 - x³)
Step 1: Reduce to Linear Form

Let y = t. Then ² y (dy)/(dx) = (dt)/(dx). The equation becomes a standard Linear Differential Equation:

(dt)/(dx) - ((2)/(x)) t = x²(2 - x³)
Step 2: Integrating Factor

Integrating Factor (I.F.):

I.F. = e∫ -(2)/(x) dx = e-2 ln x = (1)/(x²)

Solution of the LDE:

t · (I.F.) = ∫ Q(x) · (I.F.) dx + C (t)/(x²) = ∫ (1)/(x²) · x²(2 - x³) dx + C ( y)/(x²) = ∫ (2 - x³) dx + C ( y)/(x²) = 2x - (x⁴)/(4) + C
Step 3: Apply Boundary Conditions

We are given y(2) = 0 (so (0) = 0). Substitute x = 2 and y = 0:

0 = 2(2) - (2⁴)/(4) + C 0 = 4 - 4 + C C = 0

The specific solution is:

y = x² (2x - (x⁴)/(4)) = 2x³ - (x⁶)/(4)
Step 4: Evaluate tan(y(1))

Substitute x = 1:

(y(1)) = 2(1)³ - (1⁶)/(4) = 2 - (1)/(4) = (7)/(4)
Chapter Mix

Class 12 Mathematics: Differential Equations

Q14 jee_main_2026_28_january_evening Linear Differential Equations
Let y = y(x) be the solution of the differential equation x (dy)/(dx) - y = x² x, x in (0, π). If y((π)/(2))=(π)/(2), then 6y((π)/(6))-8y((π)/(4)) is equal to :
  • A. 3π
  • B. -3π
  • C. -π
  • D. π

Solution

Related Formula
d((y)/(x)) = (x dy - y dx)/(x²)
Core Logic

Rewrite the differential equation:

x dy - y dx = x² x dx

Divide by x² to create an exact differential:

(x dy - y dx)/(x²) = x dx d((y)/(x)) = x dx
Execution

Integrate both sides:

∫ d((y)/(x)) = ∫ x dx (y)/(x) = ln| x| + C

Given y((π)/(2)) = (π)/(2):

(π/2)/(π/2) = ln| (π)/(2)| + C ⇒ 1 = 0 + C ⇒ C = 1

Equation of curve: y = x(ln| x| + 1)

Evaluate at limits: y((π)/(6)) = (π)/(6)(ln(1)/(2) + 1) = (π)/(6)(-ln 2 + 1) y((π)/(4)) = (π)/(4)(ln 1√(2) + 1) = (π)/(4)(-(1)/(2)ln 2 + 1)

Calculate 6y((π)/(6)) - 8y((π)/(4)):

= 6[(π)/(6)(-ln 2 + 1)] - 8[(π)/(4)(-(1)/(2)ln 2 + 1)] = π(-ln 2 + 1) - 2π(-(1)/(2)ln 2 + 1) = -πln 2 + π + πln 2 - 2π = -π
Pattern Recognition

Whenever you see x dy - y dx, immediately test division by x², y², or xy to convert it directly into exact differential forms like d(y/x) or d(x/y).

Chapter Mix

Class 12 Maths: Differential Equations

Q52 jee_main_2025_02_april_evening Linear Differential Equations
Let f:[1,∞) → [2,∞) be a differentiable function. If 10∫₁xf(t)dt = 5xf(x) - x⁵ - 9 for all x ≥ 1, then the value of f(3) is:
  • A. 18
  • B. 32
  • C. 22
  • D. 26

Solution

Related Formula
Leibniz Rule for Differentiation under Integral Sign: (d)/(dx) ∫u(x)v(x) f(t) dt = f(v(x)) v'(x) - f(u(x)) u'(x) Standard Linear Differential Equation: (dy)/(dx) + P(x) y = Q(x)
Core Logic

Differentiating both sides with respect to x eliminates the definite integral and leads to a first-order linear differential equation.

Step 1: Differentiate both sides

Apply differentiation with respect to x using the Leibniz Rule on the left side, and product rule on the right side:

10 f(x) = 5 f(x) + 5x f'(x) - 5x⁴ 5 f(x) + 5x⁴ = 5x f'(x) f(x) + x⁴ = x f'(x)

Letting y = f(x), we rewrite it as:

(dy)/(dx) - (y)/(x) = x³
Step 2: Solve the Linear Differential Equation

The integrating factor (I.F.) is:

I.F. = e∫ -(1)/(x) dx = e-ln x = (1)/(x)

Multiply by the I.F. and integrate:

y · (1)/(x) = ∫ x³ · (1)/(x) dx = ∫ x² dx = (x³)/(3) + C

Thus, the general solution is:

f(x) = (x⁴)/(3) + C x
Step 3: Apply Boundary Conditions

Substitute x = 1 into the original integral equation:

10 ∫₁¹ f(t) dt = 5(1) f(1) - 1⁵ - 9 0 = 5 f(1) - 10 f(1) = 2

Now, substitute x = 1 and f(1) = 2 into our general solution to find C:

2 = (1)/(3) + C C = (5)/(3)

Therefore, the complete function is:

f(x) = (x⁴)/(3) + (5x)/(3)
Step 4: Compute f(3)

Evaluate the function at x = 3:

f(3) = (3⁴)/(3) + (5(3))/(3) = 27 + 5 = 32
Pattern Recognition

Whenever a definite integral is defined from a constant to the variable x within an equation, differentiating immediately reduces it to a differential equation. Finding the value of f(a) at the lower bound is a standard method to get the constant of integration.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q71 jee_main_2025_02_april_evening First Order Linear Differential Equations
Let y = y(x) be the solution of the differential equation (dy)/(dx) + 2y ² x = 2 ² x + 3 x · ² x such that y(0) = (5)/(4). Then 12(y((π)/(4)) - e⁻²) is equal to ____________.
Numerical Answer. Answer: 21 to 21

Solution

Related Formula
Linear Differential equation form: (dy)/(dx) + P(x) y = Q(x) Integrating Factor: I.F. = e∫ P(x) dx General solution: y · I.F. = ∫ Q(x) · I.F. dx + C
Core Logic

This is a first-order linear differential equation. We calculate the Integrating Factor first to write down the integral solution.

Step 1: Find the Integrating Factor (I.F.)

Here, P(x) = 2 ² x and Q(x) = 2 ² x + 3 x · ² x:

I.F. = e∫ 2 ² x dx = e2 x
Step 2: Obtain the General Solution

Multiply both sides by the integrating factor:

y · e2 x = ∫ e2 x ( 2 ² x + 3 x · ² x ) dx

Let t = x dt = ² x dx. The integral becomes:

∫ e2t (2 + 3t) dt = ∫ 2 e2t dt + 3 ∫ t e2t dt

Applying integration by parts for the second term:

3 ∫ t e2t dt = 3 [ t e2t2 - ∫ e2t2 dt ] = 3t e2t2 - 3e2t4

Summing all parts:

y · e2 x = e2t + 3t e2t2 - 3e2t4 + C = e2 x [ 1 + (3 x)/(2) - (3)/(4) ] + C y = (3 x)/(2) + (1)/(4) + C e-2 x
Step 3: Apply the boundary conditions

Using the initial boundary condition y(0) = (5)/(4):

(5)/(4) = (3(0))/(2) + (1)/(4) + C e⁰ C = 1

Thus, the complete function is:

y(x) = (3 x)/(2) + (1)/(4) + e-2 x
Step 4: Compute the final value

Evaluate the function at x = (π)/(4):

y((π)/(4)) = (3(1))/(2) + (1)/(4) + e⁻² = (7)/(4) + e⁻²

Now, calculate the requested value:

12 ( y((π)/(4)) - e⁻² ) = 12 ( (7)/(4) ) = 21
Pattern Recognition

Integration by parts substitution: When integrating terms of the form ∫ eat P(t) dt (where P(t) is a polynomial), substituting the polynomial variable directly simplifies the exponential integration factors cleanly.

Chapter Mix

Class 12 Mathematics: Differential Equations

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)