Let g be a differentiable function such that 0xg(t)dt=x-0xtg(t)dt [cite: 568], x0 [cite: 569] and let y=y(x) satisfy…

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q54 jee_main_2025_24_jan_evening Linear Differential Equations

Let

f:(0,\infty)arrow R

be a function which is differentiable at all points of its domain and satisfies the condition

x²f(x)=2xf(x)+3,

with

f(1)=4

Then

2f(2)

is equal to:

A. $29$ $29$
B. $19$ $19$
C. $39$ $39$
D. $23$ $23$

Solution

Related Formula

The quotient rule derivative identity is given by:

(d)/(dx)((f(x))/(x²)) = (x² f'(x) - 2x f(x))/(x⁴)

Core Logic

Rearrange the given differential condition:

x² f'(x) - 2x f(x) = 3

Step 1: Divide by $x⁴$ $x^4$

To convert the left-hand side into an exact derivative form, divide the full relation by $x⁴$ $x^4$:

(x² f'(x) - 2x f(x))/(x⁴) = (3)/(x⁴)

(d)/(dx)((f(x))/(x²)) = 3x⁻⁴

Step 2: Integration and Evaluating Constant

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

(f(x))/(x²) = \int 3x⁻⁴ dx = -x⁻³ + C = -(1)/(x³) + C

f(x) = -(1)/(x) + Cx²

Using the given value $f(1) = 4$ $f(1) = 4$:

4 = -(1)/(1) + C(1)² \Rightarrow 4 = -1 + C \Rightarrow C = 5

Thus, the function is $f(x) = -(1)/(x) + 5x²$ $f(x) = -\frac{1}{x} + 5x^2$.

Step 3: Calculating $2f(2)$ $2f(2)$

Substitute $x = 2$ $x = 2$ to compute $2f(2)$ $2f(2)$ :

2 \times f(2) = 2 \times [ -(1)/(2) + 5(2)² ]

2 \times f(2) = 2 \times [ -(1)/(2) + 20 ] = -1 + 40 = 39

Pattern Recognition

Recognizing the structure $x² f'(x) - 2x f(x)$ $x^2 f'(x) - 2x f(x)$ as a partial quotient rule is faster than formatting it into standard linear order $(dy)/(dx) + P(x)y = Q(x)$ $\frac{dy}{dx} + P(x)y = Q(x)$ format, though both methods lead to the identical integration parameters safely.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q73 jee_main_2025_24_jan_evening Linear Differential Equations of First Order

Let

y=y(x)

be the solution of the differential equation

2 x(dy)/(dx)= 2x-4y x, xin(0,(π)/(2))

If

y((π)/(3))=0

, then

y((π)/(4))+y((π)/(4))

is equal to \_\_\_\_.

Numerical Answer. Answer: 1

Solution

Related Formula

Standard first-order linear differential equation structure:

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

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

Step 1: Reduce into Standard Format

Divide the full expression by $2 x$ $2\cos x$:

(dy)/(dx) = (2 x x)/(2 x) - (4y x)/(2 x)

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

Step 2: Integrating Factor & Solution

Compute the integrating multiplier :

I.F. = e\int 2 x dx = e2ln| x| = ² x

Write general integration solution path :

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

y = x + C ² x

Step 3: Boundary Evaluation

Apply the initialization condition $y((π)/(3)) = 0$ $y\left(\frac{\pi}{3}\right) = 0$:

0 = ((π)/(3)) + C ²((π)/(3)) \Rightarrow 0 = (1)/(2) + C((1)/(4)) \Rightarrow C = -2

Thus, the solution is $y = x - 2 ² x$ $y = \cos x - 2\cos^2 x$ .

Find derivative $y$ $y\prime$ :

y = - x + 4 x x = - x + 2 2x

Step 4: Target Calculation

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

y((π)/(4)) = 1\sqrt(2) - 2((1)/(2)) = 1\sqrt(2) - 1

y ((π)/(4)) = - 1\sqrt(2) + 2 ((π)/(2)) = - 1\sqrt(2) + 2

y ((π)/(4)) + y((π)/(4)) = (- 1\sqrt(2) + 2) + ( 1\sqrt(2) - 1) = 1

Pattern Recognition

Linear standard layout conversions depend entirely on clear integrating factor reductions. Remember $\int x dx = ln| x|$ $\int \tan x dx = \ln|\sec x|$ clearly to safely output exact matching polynomial definitions.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q59 jee_main_2025_24_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

( xy - 5x²\sqrt(1 + x²) ) dx + (1 + x²) dy = 0

with initial condition

y(0) = 0

. Then

y(\sqrt(3))

is equal to :

A. $5\sqrt(3)2$ $\frac{5\sqrt{3}}{2}$
B. $\sqrt((14)/(3))$ $\sqrt{\frac{14}{3}}$
C. $2\sqrt(2)$ $2\sqrt{2}$
D. $\sqrt((15)/(2))$ $\sqrt{\frac{15}{2}}$

Solution

Related Formula

A linear differential equation of the first order matching $(dy)/(dx) + P(x)y = Q(x)$ $\frac{dy}{dx} + P(x)y = Q(x)$ uses an Integrating Factor written as:

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

Core Logic

Rearrange the given differential equation terms to express it in standard linear form:

(1 + x²) (dy)/(dx) + xy = 5x²\sqrt(1 + x²)

(dy)/(dx) + ((x)/(1+x²))y = 5x²\sqrt(1+x²)

Step 1: Compute Integrating Factor

Calculate the exponent integral for $I.F.$ $\text{I.F.}$:

\int P(x) dx = \int (x)/(1+x²) dx = (1)/(2) ln(1+x²) = ln\sqrt(1+x²)

I.F. = eln\sqrt(1+x²) = \sqrt(1+x²)

Step 2: General Solution and Boundary Condition

The general solution template is:

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

y\sqrt(1+x²) = \int 5x²\sqrt(1+x²) \cdot \sqrt(1+x²) dx

y\sqrt(1+x²) = \int 5x² dx = (5x³)/(3) + C

Apply the initial condition $y(0) = 0$ $y(0) = 0$:

0 \cdot \sqrt(1+0) = 0 + C C = 0

Thus, the explicit functional equation is:

y = 5x³3\sqrt(1+x²)

Step 3: Evaluate at Target Value

Substitute $x = \sqrt(3)$ $x = \sqrt{3}$ into the isolated function:

y(\sqrt(3)) = 5(\sqrt(3))³3 1+(\sqrt(3))² = 5(3\sqrt(3))3\sqrt(1+3) = 15\sqrt(3)3 \cdot 2 = 5\sqrt(3)2

Pattern Recognition

Spotting that multiplying across the differential equation format by $\sqrt(1+x²)$ $\sqrt{1+x^2}$ converts the left hand side into a direct product rule derivative matching $(d)/(dx)(y\sqrt(1+x²))$ $\frac{d}{dx}(y\sqrt{1+x^2})$ yields a direct integration pathway.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q jee_main_2025_28_jan_evening Linear Differential Equations

If

y=y(x)

is the solution of the differential equation,

4-x²(dy)/(dx)=(( ⁻¹((x)/(2)))²-y) ⁻¹((x)/(2))

-2\leq x\leq2

,

y(2)=( π²-84)

then

y²(0)

is equal to

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Standard first-order linear differential equation form:

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

Integrating factor: $I.F. = e\int P(x) dx$ $\text{I.F.} = e^{\int P(x) dx}$

Core Logic

Rearrange the given differential equation:

\sqrt(4-x²) (dy)/(dx) = ( ⁻¹((x)/(2)) )³ - y ⁻¹((x)/(2))

Divide both sides by $\sqrt(4-x²)$ $\sqrt{4-x^2}$:

(dy)/(dx) + ⁻¹(x/2)\sqrt(4-x²) y = ( ⁻¹(x/2))³\sqrt(4-x²)

Step 1: Calculate Integrating Factor and Solve

I.F. = e^\int ⁻¹(x/2)\sqrt(4-x²) dx = e^(1)/(2) ( ⁻¹(x/2))²

The general solution follows the structure:

y = ( ⁻¹((x)/(2)) )² - 2 + C \cdot e^-(1)/(2) ( ⁻¹(x/2))²

Using the initial condition $y(2) = (π² - 8)/(4) = (π²)/(4) - 2$ :

(π²)/(4) - 2 = ((π)/(2))² - 2 + C \cdot e-(π²)/(8) C = 0

Thus, the specific solution simplifies to:

y(x) = ( ⁻¹((x)/(2)) )² - 2

Step 2: Evaluate at x=0

y(0) = ( ⁻¹(0))² - 2 = 0 - 2 = -2

Therefore:

y²(0) = (-2)² = 4

Pattern Recognition

Recognizing that the coefficient of $y$ $y$ is precisely the derivative of $(1)/(2)( ⁻¹(x/2))²$ $\frac{1}{2}(\sin^{-1}(x/2))^2$ makes computing the linear integrating factor simple.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q jee_main_2025_29_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

x ( ₑ ( x))² dy + ( x - 3y x ₑ ( x)) dx = 0, x in (0, (π)/(2))

If

y((π)/(4)) = (-1)/( ₑ 2)

, then

y((π)/(6))

is:

A. $2 ₑ(3) - ₑ(4)$ $\frac{2}{\log_{e}(3) - \log_{e}(4)}$
B. $1 ₑ(4) - ₑ(3)$ $\frac{1}{\log_{e}(4) - \log_{e}(3)}$
C. $- 1 ₑ(4)$ $-\frac{1}{\log_{\mathrm{e}}(4)}$
D. $1 ₑ(3) - ₑ(4)$

Solution

Related Formula

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

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

Core Logic

Rearranging the equation to standard linear differential form:

x (ln( x))² (dy)/(dx) - 3 x ln( x)y = - x

Divide by $x (ln( x))²$ $\cos x (\ln(\cos x))^2$:

(dy)/(dx) - (3 x)/(ln( x))y = (- x)/((ln( x))²)

Alternatively, writing in terms of $x$ $\sec x$:

(dy)/(dx) + (3 x)/(ln( x))y = (- x)/((ln( x))²)

Step 1: Compute Integrating Factor

I.F. = e\int (3 x)/(ln( x)) dx = e3ln(ln( x)) = (ln( x))³

Step 2: Solve the Integral Solution

y \times (ln( x))³ = -\int ( x)/((ln( x))²) (ln( x))³ dx

y \times (ln( x))³ = -\int x ln( x) dx = -(1)/(2)(ln( x))² + C

Step 3: Apply Boundary Condition

Given $x = (π)/(4)$ $x = \frac{\pi}{4}$, $y = -(1)/(ln 2)$ $y = -\frac{1}{\ln 2}$. Note $((π)/(4)) = \sqrt(2)$ $\sec(\frac{\pi}{4}) = \sqrt{2}$.

(-(1)/(ln 2)) (ln\sqrt(2))³ = -(1)/(2)(ln\sqrt(2))² + C

(-(1)/(ln 2)) ((1)/(2)ln 2)³ = -(1)/(2)((1)/(2)ln 2)² + C C = 0

Step 4: Final Substitution for $x = π/6$ $x = \pi/6$

With $C=0$ $C=0$, $y = (-1)/(2ln( x)) = (1)/(2ln( x))$ .

y((π)/(6)) = 12ln( \sqrt(3)2) = (1)/(2((1)/(2)ln 3 - ln 2)) = (1)/(ln 3 - ln 4)

Pattern Recognition

Logarithmic functions nested inside trigonometric terms generally point to substitution structures where $ln( x)$ $\ln(\sec x)$ works cleanly alongside its derivative $x dx$ $\tan x \, dx$.

Chapter Mix

Class 12 Mathematics: Differential Equations

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Integrating Factor

Step 2: Numeric substitution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Divide by x⁴$x^4$

Step 2: Integration and Evaluating Constant

Step 3: Calculating 2f(2)$2f(2)$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Reduce into Standard Format

Step 2: Integrating Factor & Solution

Step 3: Boundary Evaluation

Step 4: Target Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Integrating Factor

Step 2: General Solution and Boundary Condition

Step 3: Evaluate at Target Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Integrating Factor and Solve

Step 2: Evaluate at x=0

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Integrating Factor

Step 2: Solve the Integral Solution

Step 3: Apply Boundary Condition

Step 4: Final Substitution for x = π/6$x = \pi/6$

Pattern Recognition

Chapter Mix

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Step 1: Divide by $x⁴$ $x^4$

Step 3: Calculating $2f(2)$ $2f(2)$

Step 4: Final Substitution for $x = π/6$ $x = \pi/6$