If for the solution curve y = f(x) of the differential equation (dy/dx) + ( x)y = (2 + x/(1 + 2 x)2), x ((-pi/2)…

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q23 jee_main_2024_27_jan_morning Reducible to Variable Separable

If the solution of the differential equation

(2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3

, is

α x+β y+3 ₑ|2x+3y-γ|=6

, then

α+2β+3γ

is equal to:

Numerical Answer. Answer: 29 to 29

Solution

Related Formula

(dy)/(dx) = f(ax+by+c)

Substitute $t = ax+by+c$ $t = ax+by+c$ to reduce the equation to variable separable form.

Core Logic

The differential equation can be written as:

(dy)/(dx) = -(2x+3y-2)/(4x+6y-7)

Observe that $4x+6y = 2(2x+3y)$ $4x+6y = 2(2x+3y)$. Let us substitute $t = 2x+3y-2$ $t = 2x+3y-2$. Taking derivatives with respect to $x$ $x$:

(dt)/(dx) = 2 + 3(dy)/(dx) \Rightarrow (dy)/(dx) = (1)/(3)((dt)/(dx) - 2)

Step 1: Translating and Simplifying

Substitute $t$ $t$ into the differential equation:

(1)/(3)((dt)/(dx) - 2) = -(t)/(2(t+2)-7)

(dt)/(dx) - 2 = -(3t)/(2t-3)

(dt)/(dx) = 2 - (3t)/(2t-3)

(dt)/(dx) = (4t - 6 - 3t)/(2t - 3) = (t - 6)/(2t - 3)

Step 2: Variable Separation Integration

Separate the variables $t$ $t$ and $x$ $x$:

\int (2t - 3)/(t - 6) dt = \int dx

Decompose the fraction algebraically:

\int (2(t-6) + 9)/(t-6) dt = \int ( 2 + (9)/(t-6) ) dt

2t + 9ln|t-6| = x + C

Step 3: Restoring Original Variables

Substitute $t = 2x + 3y - 2$ $t = 2x + 3y - 2$ back:

2(2x + 3y - 2) + 9ln|2x + 3y - 2 - 6| = x + C

4x + 6y - 4 + 9ln|2x + 3y - 8| = x + C

3x + 6y + 9ln|2x + 3y - 8| = C + 4

Divide the entire equation by 3:

x + 2y + 3ln|2x + 3y - 8| = C'

Step 4: Finding the Constant of Integration

Given initial condition $y(0) = 3$ $y(0) = 3$ (when $x=0, y=3$ $x=0, y=3$):

0 + 2(3) + 3ln|2(0) + 3(3) - 8| = C'

6 + 3ln|1| = C' \Rightarrow C' = 6

Thus, the specific solution is:

x + 2y + 3ln|2x + 3y - 8| = 6

Step 5: Comparing and Final Evaluation

Comparing with the given form $α x + β y + 3ln|2x + 3y - γ| = 6$ : $α = 1$ $\alpha = 1$, $β = 2$ $\beta = 2$, $γ = 8$ $\gamma = 8$. Compute the required expression:

α + 2β + 3γ = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29

Pattern Recognition

When the coefficients of x and y in the numerator and denominator are proportional (i.e. $a₁/a₂ = b₁/b₂$ $a_1/a_2 = b_1/b_2$), the standard procedure is to use a direct composite substitution $t = ax+by$ $t = ax+by$ which effortlessly maps to a basic logarithmic integral.

Chapter Mix

Class 12 Maths: Differential Equations

Q12 jee_main_2024_29_jan_morning Linear Differential Equations

A function

y=f(x)

satisfies

f(x) 2x+ x-(1+ ² x)f'(x)=0

with condition

f(0)=0

. Then

f((π)/(2))

is equal to

A. $1$ $1$
B. $0$ $0$
C. $-1$ $-1$
D. $2$ $2$

Solution

Related Formula

Standard Form of LDE: (dy)/(dx) + P(x)y = Q(x)

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

Solution: y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C

Core Logic

Given the differential equation:

y 2x + x - (1+ ² x)(dy)/(dx) = 0

Rearrange it into the standard linear differential equation format:

(1+ ² x)(dy)/(dx) - y 2x = x

(dy)/(dx) - (( 2x)/(1+ ² x))y = ( x)/(1+ ² x)

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

Here, $P(x) = -( 2x)/(1+ ² x)$ $P(x) = -\frac{\sin 2x}{1+\cos^2 x}$.

I.F. = e\int -( 2x)/(1+ ² x) dx

Let $1+ ² x = t$ $1+\cos^2 x = t$, then $dt = -2 x x dx = - 2x dx$ .

I.F. = e\int (1)/(t) dt = eln t = t

I.F. = 1+ ² x

Step 2: Solve the Differential Equation

Multiply the LDE by the integrating factor:

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

y(1+ ² x) = \int x dx

y(1+ ² x) = - x + C

Apply the initial condition $f(0) = 0$ $f(0) = 0$ (which means $y=0$ $y=0$ when $x=0$ $x=0$):

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

The particular solution is:

y(1+ ² x) = 1 - x

Step 3: Evaluate at x = pi/2

To find $f((π)/(2))$ $f(\frac{\pi}{2})$, substitute $x = (π)/(2)$ $x = \frac{\pi}{2}$:

y(1+ ²((π)/(2))) = 1 - ((π)/(2))

$y(1+0) = 1 - 0$ $y(1+0) = 1 - 0$ $y = 1$ $y = 1$

Pattern Recognition

Whenever $y$ $y$ and $y'$ $y'$ appear linearly separated by complex trigonometric polynomials, format aggressively into standard LDE form. The derivative of $1+ ²x$ $1+\cos^2x$ is exactly $- 2x$ $-\sin 2x$, making the logarithmic integration seamless.

Chapter Mix

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

Q26 jee_main_2024_29_jan_morning Variable Separable Form

If the solution curve

y=y(x)

of the differential equation

(1+y²)(1+ ₑ x)dx+x dy=0

,

x gt 0

passes through the point

(1, 1)

and

y(e)=(α- ((3)/(2)))/(β+ ((3)/(2)))

, then

α+2β

is

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

\int (1)/(1+y²) dy = ⁻¹y + C

\int f(x) f'(x) dx = ([f(x)]²)/(2) + C

(A - B) = ( A - B)/(1 + A B)

Core Logic

Rearrange the given differential equation to separate variables $x$ $x$ and $y$ $y$:

(1+y²)(1+ ₑ x)dx + x dy = 0

(1+ ₑ x)/(x) dx + (1)/(1+y²) dy = 0

Integrate both sides:

\int (1+ ₑ x)/(x) dx + \int (1)/(1+y²) dy = C

Step 1: Execute Integration

For the first integral, let $u = 1+ ₑ x$ $u = 1+\log_e x$, so $du = (1)/(x) dx$ $du = \frac{1}{x} dx$.

\int u du = (u²)/(2) = ((1+ ₑ x)²)/(2)

For the second integral:

\int (1)/(1+y²) dy = ⁻¹y

Putting them together:

((1+ ₑ x)²)/(2) + ⁻¹y = C

Step 2: Apply Boundary Conditions

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

((1+ ₑ 1)²)/(2) + ⁻¹(1) = C

Since $ₑ 1 = 0$ $\log_e 1 = 0$ and $⁻¹(1) = (π)/(4)$ $\tan^{-1}(1) = \frac{\pi}{4}$:

(1²)/(2) + (π)/(4) = C \Rightarrow C = (1)/(2) + (π)/(4)

Wait, the solution rearranges differently: $\int ((1)/(x) + (ln x)/(x)) dx = ln x + ((ln x)²)/(2)$ . Let's use this to stay fully synced with the PDF's algebraic step.

ln x + ((ln x)²)/(2) + ⁻¹ y = C

At $(1,1)$ $(1,1)$:

ln(1) + ((ln 1)²)/(2) + ⁻¹(1) = C \Rightarrow 0 + 0 + (π)/(4) = C \Rightarrow C = (π)/(4)

The general equation simplifies nicely to:

ln x + ((ln x)²)/(2) + ⁻¹ y = (π)/(4)

Step 3: Evaluate at x = e

Substitute $x=e$ $x=e$ into the equation to find $y(e)$ $y(e)$:

ln e + ((ln e)²)/(2) + ⁻¹ y = (π)/(4)

1 + (1²)/(2) + ⁻¹ y = (π)/(4)

(3)/(2) + ⁻¹ y = (π)/(4)

⁻¹ y = (π)/(4) - (3)/(2)

y = ((π)/(4) - (3)/(2))

Using the compound angle formula $(A-B)$ $\tan(A-B)$:

y = ( (π/4) - (3/2))/(1 + (π/4) (3/2)) = (1 - (3/2))/(1 + (3/2))

Comparing this format to $y(e) = (α- (3/2))/(β+ (3/2))$ gives:

α = 1, β = 1

Calculate $α + 2β$ $\alpha + 2\beta$: $1 + 2(1) = 3$ $1 + 2(1) = 3$

Pattern Recognition

For expressions involving separated $x$ $x$ and $y$ $y$ logarithmic terms, always split $\int (1+ln x)/(x)dx$ $\int \frac{1+\ln x}{x}dx$ into $\int (1)/(x) + \int (ln x)/(x)$ to avoid carrying a non-zero shift $(1)/(2)$ $\frac{1}{2}$ into the constant $C$ $C$. This allows direct identity matching for tangent sums.

Chapter Mix

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

Q15 jee_main_2024_30_january_evening Functional Equations

Let

f : R - 0 \to R

be a function satisfying

f((x)/(y)) = (f(x))/(f(y))

for all

x, y

,

f(y) \neq 0

. If

f'(1) = 2024

, then

A. $xf'(x) - 2024f(x) = 0$ $xf'(x) - 2024f(x) = 0$
B. $xf'(x) + 2024f(x) = 0$ $xf'(x) + 2024f(x) = 0$
C. $xf'(x) + f(x) = 2024$ $xf'(x) + f(x) = 2024$
D. $xf'(x) - 2023f(x) = 0$ $xf'(x) - 2023f(x) = 0$

Solution

Related Formula

Chain rule for partial differentiation: (\partial)/(\partial x) f((x)/(y)) = f'((x)/(y)) \cdot (1)/(y)

Core Logic

Given functional equation: $f((x)/(y)) = (f(x))/(f(y))$ First, put $x=1, y=1$ $x=1, y=1$: $f(1) = (f(1))/(f(1)) \Rightarrow f(1) = 1$ (since $f(y) \neq 0$ $f(y) \neq 0$ implies $f(1) \neq 0$ $f(1) \neq 0$).

Step 1: Partial Differentiation

Partially differentiate the given equation with respect to $x$ $x$, keeping $y$ $y$ constant:

f'((x)/(y)) \cdot (1)/(y) = (1)/(f(y)) \cdot f'(x)

Let $x \to y$ $x \to y$ so that $(x)/(y) \to 1$ $\frac{x}{y} \to 1$. Wait, let's substitute $y = x$ $y = x$ directly:

f'(1) \cdot (1)/(x) = (1)/(f(x)) \cdot f'(x)

Step 2: Forming the Final Equation

We are given $f'(1) = 2024$ $f'(1) = 2024$. Substituting this value:

2024 \cdot (1)/(x) = (f'(x))/(f(x))

Cross-multiplying yields:

2024 f(x) = x f'(x)

\Rightarrow x f'(x) - 2024 f(x) = 0

Pattern Recognition

Functional equations of the form $f(x/y) = f(x)/f(y)$ $f(x/y) = f(x)/f(y)$ inherently describe power functions ( $f(x) = x^k$ $f(x) = x^k$). Partial differentiation with respect to one variable quickly resolves the derivative form without limits.

Chapter Mix

Class 11 Maths: Functions Class 12 Maths: Differential Equations

Q21 jee_main_2024_30_january_evening Linear Differential Equation

Let

Y = Y(X)

be a curve lying in the first quadrant such that the area enclosed by the line

Y - y = Y'(x)(X - x)

and the co-ordinate axes, where

(x, y)

is any point on the curve, is always

(-y²)/(2Y'(x)) + 1

,

Y'(x) \neq 0

. If

Y(1) = 1

, then

12Y(2)

equals

Numerical Answer. Answer: 20 to 20

Solution

Related Formula

Equation of tangent at (x, y): Y - y = (dy)/(dx)(X - x)

Area of Right Triangle: (1)/(2) \times base \times height

Core Logic

The line $Y - y = y'(X - x)$ $Y - y = y'(X - x)$ intersects the axes to form a triangle. Let's find the intercepts: X-intercept (set $Y=0$ $Y=0$): $-y = y'(X - x) \Rightarrow X = x - (y)/(y')$ Y-intercept (set $X=0$ $X=0$): $Y - y = y'(-x) \Rightarrow Y = y - xy'$

The area of the triangle bounded by the coordinate axes is:

Area = (1)/(2) ( x - (y)/(y') ) (y - xy') = (-y²)/(2y') + 1

Linear Differential Equation diagram for Q21 - JEE Main 2024 Evening

Step 1: Setting up the Differential Equation

Expand the Area equation:

(1)/(2) [ xy - x²y' - (y²)/(y') + xy ] = (-y²)/(2y') + 1

Multiply by 2:

2xy - x²y' - (y²)/(y') = -(y²)/(y') + 2

Cancel $-(y²)/(y')$ $-\frac{y^2}{y'}$ from both sides: $2xy - x²y' = 2$ $2xy - x^2y' = 2$

Step 2: Solving the Linear Differential Equation

Rearrange into standard LDE form:

x² (dy)/(dx) - 2xy = -2

(dy)/(dx) - (2)/(x)y = -(2)/(x²)

Find the Integrating Factor (IF):

I.F. = e\int -(2)/(x) dx = e-2ln x = x⁻² = (1)/(x²)

Multiply the LDE by the IF:

y \cdot (1)/(x²) = \int ( -(2)/(x²) ) (1)/(x²) dx + C

y x⁻² = -2 \int x⁻⁴ dx + C

y x⁻² = -2 ( x⁻³-3 ) + C

(y)/(x²) = (2)/(3x³) + C

y = (2)/(3x) + Cx²

Step 3: Applying Initial Conditions

Given $Y(1) = 1$ $Y(1) = 1$ (i.e., when $x=1, y=1$ $x=1, y=1$):

1 = (2)/(3(1)) + C(1)² \Rightarrow 1 = (2)/(3) + C \Rightarrow C = (1)/(3)

Thus, the curve is:

y = (2)/(3x) + (x²)/(3)

We need to find $12Y(2)$ $12Y(2)$:

Y(2) = (2)/(3(2)) + (2²)/(3) = (1)/(3) + (4)/(3) = (5)/(3)

12Y(2) = 12 ( (5)/(3) ) = 4 \times 5 = 20

Pattern Recognition

Geometrical word problems forming an Area differential equation almost always reduce to a first-order Linear Differential Equation (LDE) or an Exact differential form once you cleanly substitute the axis intercepts.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Application of Derivatives

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Integration with Half-Angle Substitutions

Step 2: Calculate Target Point Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Translating and Simplifying

Step 2: Variable Separation Integration

Step 3: Restoring Original Variables

Step 4: Finding the Constant of Integration

Step 5: Comparing and Final Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

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

Step 2: Solve the Differential Equation

Step 3: Evaluate at x = pi/2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Execute Integration

Step 2: Apply Boundary Conditions

Step 3: Evaluate at x = e

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Partial Differentiation

Step 2: Forming the Final Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Differential Equation

Step 2: Solving the Linear Differential Equation

Step 3: Applying Initial Conditions

Pattern Recognition

Chapter Mix

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