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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 4

Q72 jee_main_2025_02_april_morning Solving First Order Differential Equations
Let f: R → R be a thrice differentiable odd function satisfying f'(x) ≥ 0, f'(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f( ₑ 3) is equal to ________.
Numerical Answer. Answer: 36 to 36

Solution

Related Formula

Standard variable separable integration form:

∫ 1√(y² + a²) dy = ln|y + √(y² + a²)| + C
Core Logic

The original paper solution states the structure equation setup as f''(x) = f(x). Multiply by f'(x) on both sides to transform it into a integrable derivative form.

Step 1: Integrate the derivative identity
f'(x) · f''(x) = f'(x) · f(x)

Integrate both sides with respect to x:

((f'(x))²)/(2) = ((f(x))²)/(2) + C (f'(x))² = (f(x))² + C'
Step 2: Find the constant of integration

Use initial conditions f(0) = 0 and f'(0) = 3:

3² = 0² + C' C' = 9

Thus, (f'(x))² = (f(x))² + 9. Given f'(x) ≥ 0:

f'(x) = √((f(x))² + 9)
Step 3: Variable Separation and Solution Form

Let y = f(x) dydx = √(y² + 9):

∫ dy√(y² + 9) = ∫ dx ln|y + √(y² + 9)| = x + C₂

Substitute initial condition x=0, y=0:

ln|0 + √(9)| = 0 + C₂ C₂ = ln 3

Therefore, ln|y + √(y² + 9)| = x + ln 3 y + √(y² + 9) = 3e^x.

Step 4: Compute targeted value

We need to evaluate at x = ln 3:

y + √(y² + 9) = 3eln 3 = 3(3) = 9 √(y² + 9) = 9 - y

Square both sides:

y² + 9 = 81 - 18y + y² 18y = 72 y = 4

Thus, f(ln 3) = 4. Multiply by 9:

9 f(ln 3) = 9(4) = 36
Pattern Recognition

Multiplying a second derivative by the first derivative (f'f'') is a classic trick to convert a second-order linear differential equation into a first-order separable layout, opening a clear path to the solution.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Differential Calculus

Q56 jee_main_2025_07_april_morning Linear Differential Equations
Let y = y(x) be the solution curve of the differential equation x(x² + e^x)dy + (e^x(x - 2)y - x³)dx = 0, x > 0, passing through the point (1,0) . Then y(2) is equal to:
  • A. 44 - e²
  • B. 22 + e²
  • C. 22 - e²
  • D. 44 + e²

Solution

Related Formula

Standard first-order linear differential equation form:

(dy)/(dx) + P(x)y = Q(x)

Integrating Factor:

I.F. = e∫ P(x)dx

Solution layout:

y · (I.F.) = ∫ Q(x) · (I.F.) dx + C
Core Logic

Rearrange the given differential equation into standard linear form:

x(x² + e^x)(dy)/(dx) + e^x(x - 2)y = x³ (dy)/(dx) + (e^x(x - 2))/(x(x² + e^x))y = (x²)/(x² + e^x)
Step 1: Determine Integrating Factor

We need to integrate P(x) = (e^x(x-2))/(x(x²+e^x)):

∫ (e^x(x-2))/(x(x²+e^x)) dx = ∫ (xe^x - 2e^x)/(x(x²+e^x)) dx = ∫ ((xe^x - 2e^x)/(x³))/(1 + (e^x)/(x²)) dx

Let t = 1 + (e^x)/(x²). Then dt = (x²e^x - e^x(2x))/(x⁴) dx = (xe^x - 2e^x)/(x³) dx.

Thus:

∫ P(x)dx = ∫ (dt)/(t) = ln|t| = ln|1 + (e^x)/(x²)|

Therefore, the integrating factor is:

I.F. = eln(1 + (e^x)/(x²)) = 1 + (e^x)/(x²) = (x² + e^x)/(x²)
Step 2: Construct the General Solution

Using the linear equation solution template:

y · ((x² + e^x)/(x²)) = ∫ ((x²)/(x² + e^x)) · ((x² + e^x)/(x²)) dx + C y · (1 + (e^x)/(x²)) = ∫ 1 · dx + C y · (1 + (e^x)/(x²)) = x + C
Step 3: Evaluate Constant and Compute y(2)

The curve passes through (1, 0). Substitute x=1, y=0:

0 · (1 + e) = 1 + C C = -1

So the exact solution equation is:

y · (1 + (e^x)/(x²)) = x - 1 y = (x - 1)/(1 + (e^x)/(x²))

To find y(2), substitute x=2:

y(2) = (2 - 1)/(1 + (e²)/(2²)) = (1)/(1 + (e²)/(4)) = (4)/(4 + e²)
Pattern Recognition

Spotting the functional derivative framework inside P(x) by dividing the numerator and denominator by x³ uncovers the standard form ∫ (f'(x))/(f(x))dx cleanly, converting an otherwise intimidating integral into a basic natural log operation.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integrals

Q jee_main_2025_08_april_evening Linear Differential Equations
Let f(x) = x - 1 and g(x) = e^x for x in R. If (dy)/(dx) = (e-2√(x) g(f(f(x))) - y√(x)), y(0) = 0, then y(1) is:
  • A. 1 - e²e⁴
  • B. 2e - 1e³
  • C. e - 1e⁴
  • D. 1 - e³e⁴

Solution

Related Formula
Linear Form: (dy)/(dx) + P(x)y = Q(x) I.F. = e∫ P(x) dx
Core Logic

Evaluate composite function layers to organize equation segments into a standard first-order linear differential form, then introduce proper scaling factors.

Step 1: Simplify Composite Functional Core
f(f(x)) = (x-1) - 1 = x - 2 g(f(f(x))) = ex-2
Step 2: Restructure Equation and Compute Integrating Factor
(dy)/(dx) + 1√(x)y = e-2√(x) · ex-2 = ex - 2√(x) - 2 I.F. = e^∫ 1√(x) dx = e2√(x)
Step 3: Integrate General Tracking Steps
y × e2√(x) = ∫ e2√(x) × ex - 2√(x) - 2 dx + c = ∫ ex-2 dx + c y × e2√(x) = ex-2 + c

Using boundary values x=0, y=0 0 = e⁻² + c c = -e⁻².

Step 4: Evaluate Value Bounds At Point Profile
y × e2√(x) = ex-2 - e⁻²

At x = 1:

y(1) × e² = e⁻¹ - e⁻² y(1) = e⁻¹ - e⁻²e² = (e-1)/(e⁴)
Pattern Recognition

Composite layouts often produce exponent segments designed to cancel tracking multiples within integrating factor components automatically.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 11 Mathematics: Relations and Functions

Q70 jee_main_2025_29_jan_evening Linear Differential Equations
If for the solution curve y = f(x) of the differential equation (dy)/(dx) + ( x)y = (2 + x)/((1 + 2 x)²), x in ((-π)/(2), (π)/(2)), f((π)/(3)) = √(3)10, then f((π)/(4)) is equal to:
  • A. 9√(3) + 310(4 + √(3))
  • B. √(3) + 110(4 + √(3))
  • C. 5 - √(3)2√(2)
  • D. 4 - √(2)14

Solution

Related Formula

Integrating factor (I.F.) for a linear differential equation (dy)/(dx) + Py = Q:

I.F. = e∫ P dx

General solution:

y · (I.F.) = ∫ Q · (I.F.) dx
Core Logic

Given P = x, compute Integrating Factor:

I.F. = e∫ x dx = eln( x) = x

Set up integrated expression solution layout:

y · x = ∫ (2 + x)/((1 + 2 x)²) · x dx = ∫ (2 x + 1)/(( x + 2)²) · dx
Step 1: Evaluate Integration with Half-Angle Substitutions

Using tangent half-angle substitution t = (x)/(2) transformations simplifies the integral loop structure down to:

y · x = (2)/(t + (3)/(t)) + C

Plugging entry condition parameters f((π)/(3)) = √(3)10 tracking t = 1√(3) explicitly isolates boundary condition constant C: C = 0

Step 2: Calculate Target Point Value

At target query point x = (π)/(4), half-angle parameters scale to t = √(2) - 1:

y · √(2) = 2√(2) - 1 + 3√(2) - 1 = 2(√(2) - 1)6 - 2√(2) y = 4 - √(2)14
Pattern Recognition

When integrating complex rational expressions involving trigonometric values, half-angle substitution methods (t = (x)/(2)) are standard for reducing polynomial degrees.

Chapter Mix

Class 12 Mathematics: Differential Equations

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