Let $f:(0,infty)rightarrow R$ $f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2f^prime(x)=2xf(x)+3,$ $x^{2}f^{\prime}(x)=2xf(x)+3,$ with $f(1)=4$ $f(1)=4$ Then $2f(2)$ $2f(2)$ is equal to: [cite: 3263, 3264]

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

Solution & Explanation

### Related Formula The quotient rule derivative identity is given by:

fracddxleft(fracf(x)x^2right) = fracx^2 f'(x) - 2x f(x)x^4

### Core Logic Rearrange the given differential condition:

x^2 f'(x) - 2x f(x) = 3

### Step 1: Divide by

x^4

To convert the left-hand side into an exact derivative form, divide the full relation by

x^4

fracx^2 f'(x) - 2x f(x)x^4 = frac3x^4

fracddxleft(fracf(x)x^2right) = 3x^-4

### Step 2: Integration and Evaluating Constant Integrating both sides with respect to

x

[cite: 3893, 3894]:

fracf(x)x^2 = int 3x^-4 dx = -x^-3 + C = -frac1x^3 + C

f(x) = -frac1x + Cx^2

Using the given value

f(1) = 4

[cite: 3264, 3896]:

4 = -frac11 + C(1)^2 Rightarrow 4 = -1 + C Rightarrow C = 5

Thus, the function is

f(x) = -frac1x + 5x^2

. ### Step 3: Calculating

2f(2) Substitute

x = 2

to compute

2f(2)

:

2 times f(2) = 2 times left[ -frac12 + 5(2)^2 right]

2 times f(2) = 2 times left[ -frac12 + 20 right] = -1 + 40 = 39

[cite: 3900, 3901] ### Pattern Recognition Recognizing the structure

x^2 f'(x) - 2x f(x)

as a partial quotient rule is faster than formatting it into standard linear order

\frac{dy}{dx} + P(x)y = Q(x)$ format, though both methods lead to the identical integration parameters safely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Reference Study Guides

Differential Equations

More Differential Equations Previous-Year Questions — Page 4

Q75 2025 Linear Differential Equations

y=y(x)

is the solution of the differential equation,

sqrt4-x^2fracdydx=left(left(sin^-1left(fracx2right)right)^2-yright)sin^-1left(fracx2right)

-2le xle2

y(2)=left(fracpi^2-84right)

then

y^2(0)

is equal to

Numerical Answer. Answer: 4 to 4

Solution

### Related Formula Standard first-order linear differential equation form:

fracdydx + P(x)y = Q(x)

Integrating factor:

textI.F. = e^int P(x) dx

### Core Logic Rearrange the given differential equation:

sqrt4-x^2 fracdydx = left( sin^-1left(fracx2right) right)^3 - y sin^-1left(fracx2 ight)

Divide both sides by

sqrt4-x^2

fracdydx + fracsin^-1(x/2)sqrt4-x^2 y = frac(sin^-1(x/2))^3sqrt4-x^2

### Step 1: Calculate Integrating Factor and Solve

textI.F. = e^int fracsin^-1(x/2)sqrt4-x^2 dx = e^frac12 (sin^-1(x/2))^2

The general solution follows the structure:

y = left( sin^-1left(fracx2right) right)^2 - 2 + C cdot e^-frac12 (sin^-1(x/2))^2

Using the initial condition

y(2) = fracpi^2 - 84 = fracpi^24 - 2

fracpi^24 - 2 = left(fracpi2right)^2 - 2 + C cdot e^-fracpi^28 implies C = 0

Thus, the specific solution simplifies to:

y(x) = left( sin^-1left(fracx2right) right)^2 - 2

### Step 2: Evaluate at x=0

$y(0) = (sin^-1(0))^2 - 2 = 0 - 2 = -2

Therefore:

y^2(0) = (-2)^2 = 4

### Pattern Recognition Recognizing that the coefficient of

y

is precisely the derivative of

frac12(sin^-1(x/2))^2

makes computing the linear integrating factor simple. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Q60 2025 Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

cos x (log_e(cos x))^2 \, dy + left(sin x - 3y sin x log_e(cos x)right) d x = 0, x in left(0, fracpi2right).

yleft(fracpi4 ight) = frac-1log_e 2

, then

yleft(fracpi6 ight)

is:

A. $frac2log_e(3) - log_e(4)$ $\frac{2}{\log_{e}(3) - \log_{e}(4)}$
B. $frac1log_e(4) - log_e(3)$ $\frac{1}{\log_{e}(4) - \log_{e}(3)}$
C. $-frac1log_mathrme(4)$ $-\frac{1}{\log_{\mathrm{e}}(4)}$
D. $frac1log_mathrme(3) - log_mathrme(4)$

Solution

### Related Formula

textStandard Linear Form: fracdydx + P(x)y = Q(x)

textIntegrating Factor (I.F.) = e^int P(x) \, dx

### Core Logic Rearranging the equation to standard linear differential form:

cos x (ln(cos x))^2 fracdydx - 3sin x ln(cos x)y = -sin x

Divide by

cos x (ln(cos x))^2

fracdydx - frac3tan xln(cos x)y = frac-tan x(ln(cos x))^2

Alternatively, writing in terms of

sec x

fracdydx + frac3tan xln(sec x)y = frac-tan x(ln(sec x))^2

### Step 1: Compute Integrating Factor

I.F. = e^int frac3tan xln(sec x) \, dx = e^3ln(ln(sec x)) = (ln(sec x))^3

### Step 2: Solve the Integral Solution

y times (ln(sec x))^3 = -int fractan x(ln(sec x))^2 (ln(sec x))^3 \, dx

y times (ln(sec x))^3 = -int tan x ln(sec x) \, dx = -frac12(ln(sec x))^2 + C

### Step 3: Apply Boundary Condition Given

x = fracpi4

y = -frac1ln 2

. Note

sec(fracpi4) = sqrt2

left(-frac1ln 2right) left(lnsqrt2right)^3 = -frac12left(lnsqrt2right)^2 + C

left(-frac1ln 2right) left(frac12ln 2right)^3 = -frac12left(frac12ln 2right)^2 + C implies C = 0

### Step 4: Final Substitution for

x = pi/6

With

C=0

y = frac-12ln(sec x) = frac12ln(cos x)

yleft(fracpi6right) = frac12lnleft(fracsqrt32right) = frac12left(frac12ln 3 - ln 2right) = frac1ln 3 - ln 4

### Pattern Recognition Logarithmic functions nested inside trigonometric terms generally point to substitution structures where

ln(sec x)

works cleanly alongside its derivative

tan x \, dx

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

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Let $f:(0,infty)rightarrow R$ $f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2f^prime(x)=2xf(x)+3,$ $x^{2}f^{\prime}(x)=2xf(x)+3,$ with $f(1)=4$ $f(1)=4$ Then $2f(2)$ $2f(2)$ is equal to: [cite: 3263, 3264]

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 4

Solution

Solution

More Differential Equations Questions — jee_main_2025_24_jan_evening

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Let f:(0,infty)rightarrow R$f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition x^2f^prime(x)=2xf(x)+3,$x^{2}f^{\prime}(x)=2xf(x)+3,$ with f(1)=4$f(1)=4$ Then 2f(2)$2f(2)$ is equal to: [cite: 3263, 3264]

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 4

Solution

Solution

More Differential Equations Questions — jee_main_2025_24_jan_evening

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Let $f:(0,infty)rightarrow R$ $f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2f^prime(x)=2xf(x)+3,$ $x^{2}f^{\prime}(x)=2xf(x)+3,$ with $f(1)=4$ $f(1)=4$ Then $2f(2)$ $2f(2)$ is equal to: [cite: 3263, 3264]