Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $left( xy - 5x^2sqrt1 + x^2 right) dx + (1 + x^2) dy = 0$ with initial condition $y(0) = 0$ $y(0) = 0$ . Then $y(sqrt3)$ $y(\sqrt{3})$ is equal to :

A. $frac5sqrt32$ $\frac{5\sqrt{3}}{2}$
B. $sqrtfrac143$ $\sqrt{\frac{14}{3}}$
C. $2sqrt2$ $2\sqrt{2}$
D. $sqrtfrac152$ $\sqrt{\frac{15}{2}}$

Solution & Explanation

### Related Formula A linear differential equation of the first order matching

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

uses an Integrating Factor written as:

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

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

(1 + x^2) fracdydx + xy = 5x^2sqrt1 + x^2

fracdydx + left(fracx1+x^2right)y = frac5x^2sqrt1+x^2

### Step 1: Compute Integrating Factor Calculate the exponent integral for

textI.F.

int P(x) dx = int fracx1+x^2 dx = frac12 ln(1+x^2) = lnsqrt1+x^2

textI.F. = e^lnsqrt1+x^2 = sqrt1+x^2

### Step 2: General Solution and Boundary Condition The general solution template is:

y cdot (textI.F.) = int Q(x) cdot (textI.F.) dx

ysqrt1+x^2 = int frac5x^2sqrt1+x^2 cdot sqrt1+x^2 dx

ysqrt1+x^2 = int 5x^2 dx = frac5x^33 + C

Apply the initial condition

y(0) = 0

0 cdot sqrt1+0 = 0 + C implies C = 0

Thus, the explicit functional equation is:

y = frac5x^33sqrt1+x^2

### Step 3: Evaluate at Target Value Substitute

x = sqrt3

into the isolated function:

y(sqrt3) = frac5(sqrt3)^33sqrt1+(sqrt3)^2 = frac5(3sqrt3)3sqrt1+3 = frac15sqrt33 cdot 2 = frac5sqrt32

### Pattern Recognition Spotting that multiplying across the differential equation format by

sqrt1+x^2

converts the left hand side into a direct product rule derivative matching

fracddx(ysqrt1+x^2)

yields a direct integration pathway. ### 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 $y = y(x)$ $y = y(x)$ be the solution of the differential equation $left( xy - 5x^2sqrt1 + x^2 right) dx + (1 + x^2) dy = 0$ with initial condition $y(0) = 0$ $y(0) = 0$ . Then $y(sqrt3)$ $y(\sqrt{3})$ is equal to :

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 4

Solution

Solution

More Differential Equations Questions — jee_main_2025_24_jan_morning

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Let y = y(x)$y = y(x)$ be the solution of the differential equation left( xy - 5x^2sqrt1 + x^2 right) dx + (1 + x^2) dy = 0$\left( xy - 5x^2\sqrt{1 + x^2} \right) dx + (1 + x^2) dy = 0$ with initial condition y(0) = 0$y(0) = 0$ . Then y(sqrt3)$y(\sqrt{3})$ is equal to :

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 4

Solution

Solution

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Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $left( xy - 5x^2sqrt1 + x^2 right) dx + (1 + x^2) dy = 0$ with initial condition $y(0) = 0$ $y(0) = 0$ . Then $y(sqrt3)$ $y(\sqrt{3})$ is equal to :