Let y = y(x) be the solution of the differential equation (dy/dx) = 2x(x+y)3 - x(x+y)

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q72 jee_main_2025_02_april_morning Solving First Order Differential Equations

Let

f: R \to R

be a thrice differentiable odd function satisfying

f'(x) \geq 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:

\int 1\sqrt(y² + a²) dy = ln|y + \sqrt(y² + a²)| + C

Core Logic

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

Step 1: Integrate the derivative identity

f'(x) \cdot f''(x) = f'(x) \cdot f(x)

Integrate both sides with respect to $x$ $x$:

((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 $f(0) = 0$ $f(0) = 0$ and $f'(0) = 3$ $f'(0) = 3$:

3² = 0² + C' C' = 9

Thus, $(f'(x))² = (f(x))² + 9$ $(f'(x))^2 = (f(x))^2 + 9$. Given $f'(x) \geq 0$ $f'(x) \ge 0$:

f'(x) = \sqrt((f(x))² + 9)

Step 3: Variable Separation and Solution Form

Let $y = f(x) dydx = \sqrt(y² + 9)$ :

\int dy\sqrt(y² + 9) = \int dx ln|y + \sqrt(y² + 9)| = x + C₂

Substitute initial condition $x=0, y=0$ $x=0, y=0$:

ln|0 + \sqrt(9)| = 0 + C₂ C₂ = ln 3

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

Step 4: Compute targeted value

We need to evaluate at $x = ln 3$ $x = \ln 3$:

y + \sqrt(y² + 9) = 3eln 3 = 3(3) = 9

\sqrt(y² + 9) = 9 - y

Square both sides:

y² + 9 = 81 - 18y + y² 18y = 72 y = 4

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

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

Pattern Recognition

Multiplying a second derivative by the first derivative ( $f'f''$ $f'f''$) 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

Q jee_main_2025_03_april_evening Linear Differential Equations

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

A. $(2)/(3)$ $\frac{2}{3}$
B. $(4)/(3)$ $\frac{4}{3}$
C. $(4)/(3) + e³$ $\frac{4}{3} + e^3$
D. $(2)/(3) + e³$ $\frac{2}{3} + e^3$

Solution

Related Formula

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

Integrating Factor (I.F.) = $e\int P(x) dx$ $e^{\int P(x) \, dx}$
Solution is $y \cdot I.F. = \int Q(x) \cdot I.F. dx + C$

Core Logic

Let's simplify the coefficient of $y$ $y$:

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. = $e\int 3 ² x dx = e3 x$

The general solution is:

y \cdot e3 x = \int ² x \cdot e3 x dx + C

Substitute $u = 3 x du = 3 ² x dx$ :

y \cdot e3 x = (1)/(3) \int e^u du + C = (1)/(3) e3 x + C

Step 2: Solving for boundary conditions

Given $y(0) = (1)/(3) + e³$ $y(0) = \frac{1}{3} + e^3$:

((1)/(3) + e³) \cdot e⁰ = (1)/(3) e⁰ + C C = e³

Thus, the explicit function is:

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

Evaluating at $x = (π)/(4)$ $x = \frac{\pi}{4}$:

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

Pattern Recognition

Recognizing that $3 ² x + 3 = 3 ² x$ $3\tan^2 x + 3 = 3\sec^2 x$ converts the system immediately into a classic linear differential equation where the coefficient of $y$ $y$ 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

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. $44 - e²$ $\frac{4}{4 - \mathrm{e}^{2}}$
B. $22 + e²$ $\frac{2}{2 + \mathrm{e}^2}$
C. $22 - e²$ $\frac{2}{2 - \mathrm{e}^{2}}$
D. $44 + e²$ $\frac{4}{4 + \mathrm{e}^{2}}$

Solution

Related Formula

Standard first-order linear differential equation form:

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

Integrating Factor:

I.F. = e\int P(x)dx

Solution layout:

y \cdot (I.F.) = \int Q(x) \cdot (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 $P(x) = (e^x(x-2))/(x(x²+e^x))$ $P(x) = \frac{e^x(x-2)}{x(x^2+e^x)}$:

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

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

Thus:

\int P(x)dx = \int (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 \cdot ((x² + e^x)/(x²)) = \int ((x²)/(x² + e^x)) \cdot ((x² + e^x)/(x²)) dx + C

y \cdot (1 + (e^x)/(x²)) = \int 1 \cdot dx + C

y \cdot (1 + (e^x)/(x²)) = x + C

Step 3: Evaluate Constant and Compute y(2)

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

0 \cdot (1 + e) = 1 + C C = -1

So the exact solution equation is:

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

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

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

Pattern Recognition

Spotting the functional derivative framework inside $P(x)$ $P(x)$ by dividing the numerator and denominator by $x³$ $x^3$ uncovers the standard form $\int (f'(x))/(f(x))dx$ $\int \frac{f'(x)}{f(x)}dx$ 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\sqrt(x) g(f(f(x))) - y\sqrt(x))

,

y(0) = 0

, then

y(1)

is:

A. $1 - e²e⁴$ $\frac{1 - \mathrm{e}^2}{\mathrm{e}^4}$
B. $2e - 1e³$ $\frac{2\mathrm{e} - 1}{\mathrm{e}^3}$
C. $e - 1e⁴$ $\frac{\mathrm{e} - 1}{\mathrm{e}^4}$
D. $1 - e³e⁴$ $\frac{1 - \mathrm{e}^3}{\mathrm{e}^4}$

Solution

Related Formula

Linear Form: (dy)/(dx) + P(x)y = Q(x) I.F. = e\int 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\sqrt(x)y = e-2\sqrt(x) \cdot ex-2 = ex - 2\sqrt(x) - 2

I.F. = e^\int 1\sqrt(x) dx = e2\sqrt(x)

Step 3: Integrate General Tracking Steps

y \times e2\sqrt(x) = \int e2\sqrt(x) \times ex - 2\sqrt(x) - 2 dx + c = \int ex-2 dx + c

y \times e2\sqrt(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 \times e2\sqrt(x) = ex-2 - e⁻²

At $x = 1$ $x = 1$:

y(1) \times 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)) = \sqrt(3)10

, then

f((π)/(4))

is equal to:

A. $9\sqrt(3) + 310(4 + \sqrt(3))$
B. $\sqrt(3) + 110(4 + \sqrt(3))$
C. $5 - \sqrt(3)2\sqrt(2)$ $\frac{5 - \sqrt{3}}{2\sqrt{2}}$
D. $4 - \sqrt(2)14$ $\frac{4 - \sqrt{2}}{14}$

Solution

Related Formula

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

I.F. = e\int P dx

General solution:

y \cdot (I.F.) = \int Q \cdot (I.F.) dx

Core Logic

Given $P = x$ $P = \tan x$, compute Integrating Factor:

I.F. = e\int x dx = eln( x) = x

Set up integrated expression solution layout:

y \cdot x = \int (2 + x)/((1 + 2 x)²) \cdot x dx = \int (2 x + 1)/(( x + 2)²) \cdot dx

Step 1: Evaluate Integration with Half-Angle Substitutions

Using tangent half-angle substitution $t = (x)/(2)$ $t = \tan\frac{x}{2}$ transformations simplifies the integral loop structure down to:

y \cdot x = (2)/(t + (3)/(t)) + C

Plugging entry condition parameters $f((π)/(3)) = \sqrt(3)10$ tracking $t = 1\sqrt(3)$ $t = \frac{1}{\sqrt{3}}$ explicitly isolates boundary condition constant $C$ $C$: $C = 0$ $C = 0$

Step 2: Calculate Target Point Value

At target query point $x = (π)/(4)$ $x = \frac{\pi}{4}$, half-angle parameters scale to $t = \sqrt(2) - 1$ $t = \sqrt{2} - 1$:

y \cdot \sqrt(2) = 2\sqrt(2) - 1 + 3\sqrt(2) - 1 = 2(\sqrt(2) - 1)6 - 2\sqrt(2)

y = 4 - \sqrt(2)14

Pattern Recognition

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

Chapter Mix

Class 12 Mathematics: Differential Equations

Let y = y(x)$y = y(x)$ be the solution of the differential equation (dy)/(dx) = 2x(x+y)³ - x(x+y) - 1$\frac{dy}{dx} = 2x(x+y)^{3} - x(x+y) - 1$, with y(0) = 1$y(0) = 1$. Then, ( 1√(2) + y( 1√(2)) )²$\left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^{2}$ equals:

Solution & Explanation

Related Formula

Core Logic

Step 1: Separating Variables and Integrating

Step 2: Apply the Boundary Condition

Step 3: Evaluate at the Target Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Integrate the derivative identity

Step 2: Find the constant of integration

Step 3: Variable Separation and Solution Form

Step 4: Compute targeted value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Integrating Factor and General Solution

Step 2: Solving for boundary conditions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Integrating Factor

Step 2: Construct the General Solution

Step 3: Evaluate Constant and Compute y(2)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Composite Functional Core

Step 2: Restructure Equation and Compute Integrating Factor

Step 3: Integrate General Tracking Steps

Step 4: Evaluate Value Bounds At Point Profile

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate Integration with Half-Angle Substitutions

Step 2: Calculate Target Point Value

Pattern Recognition

Chapter Mix

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Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $(dy)/(dx) = 2x(x+y)³ - x(x+y) - 1$ , with $y(0) = 1$ $y(0) = 1$. Then, $( 1\sqrt(2) + y( 1\sqrt(2)) )²$ equals: