Solution & Explanation

### Related Formula For a linear differential equation of the form

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

textIntegrating Factor (IF) = e^int P(x)dx

textSolution is y cdot textIF = int (Q(x) cdot textIF) dx + C

### Core Logic Rearrange the given differential equation to standard linear form:

(1 + x^2)dy = (tan^-1x - y)dx

fracdydx + fracyx^2 + 1 = fractan^-1xx^2 + 1

### Step 1: Find the Integrating Factor

P(x) = frac1x^2 + 1

textIF = e^int frac1x^2 + 1 dx = e^tan^-1x

### Step 2: Solve the Integral

y cdot e^tan^-1x = int e^tan^-1x cdot fractan^-1x1 + x^2 dx

Let

t = tan^-1x

, then

dt = frac11 + x^2 dx

. The integral becomes

int t e^t dt

. Using integration by parts:

int t e^t dt = t e^t - e^t + C

Substituting back

t = tan^-1x

y cdot e^tan^-1x = tan^-1x(e^tan^-1x) - e^tan^-1x + C

### Step 3: Apply Boundary Condition Given

y(0) = 1

1 cdot e^0 = 0 cdot e^0 - e^0 + C

1 = 0 - 1 + C Rightarrow C = 2

### Step 4: Evaluate at x = 1 Equation of curve:

y = tan^-1x - 1 + 2e^-tan^-1x

Evaluate at

x = 1

y(1) = tan^-1(1) - 1 + 2e^-tan^-1(1)

y(1) = fracpi4 - 1 + 2e^-pi/4

y(1) = frac2e^pi/4 + fracpi4 - 1

### Pattern Recognition A classic LDE integration trap:

int e^f(x) f(x) f'(x) dx

resolves trivially with the substitution

u = f(x)

turning it into

int u e^u du

, which always evaluates to

e^u(u - 1) + C

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

More Differential Equations Previous-Year Questions — Page 4

Q73 jee_main_2025_24_jan_evening Linear Differential Equations of First Order

Let

y=y(x)

be the solution of the differential equation

2cos xfracdydx=sin 2x-4ysin x, xinleft(0,fracpi2right)

[cite: 3407, 3411] If

yleft(fracpi3right)=0

, then

y^primeleft(fracpi4right)+yleft(fracpi4right)

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

Numerical Answer. Answer: 1

Solution

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

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

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

### Step 1: Reduce into Standard Format Divide the full expression by

2cos x

[cite: 3411, 4085]:

fracdydx = frac2sin x cos x2cos x - frac4ysin x2cos x

fracdydx + 2ytan x = sin x

### Step 2: Integrating Factor & Solution Compute the integrating multiplier :

textI.F. = e^int 2tan x dx = e^2ln|sec x| = sec^2 x

Write general integration solution path :

y cdot sec^2 x = int sin x cdot sec^2 x dx = int sec x tan x dx = sec x + C

y = cos x + Ccos^2 x

### Step 3: Boundary Evaluation Apply the initialization condition

yleft(fracpi3right) = 0

[cite: 3412, 4087]:

0 = cosleft(fracpi3right) + Ccos^2left(fracpi3right) Rightarrow 0 = frac12 + Cleft(frac14right) Rightarrow C = -2

Thus, the solution is

y = cos x - 2cos^2 x

. Find derivative

yprime

yprime = -sin x + 4cos x sin x = -sin x + 2sin 2x

### Step 4: Target Calculation Evaluate components at

x = fracpi4

yleft(fracpi4right) = frac1sqrt2 - 2left(frac12right) = frac1sqrt2 - 1

yprimeleft(fracpi4right) = -frac1sqrt2 + 2sinleft(fracpi2right) = -frac1sqrt2 + 2

yprimeleft(fracpi4right) + yleft(fracpi4right) = left(-frac1sqrt2 + 2right) + left(frac1sqrt2 - 1right) = 1

### Pattern Recognition Linear standard layout conversions depend entirely on clear integrating factor reductions. Remember

int tan x dx = ln|sec x|

clearly to safely output exact matching polynomial definitions. ### Evaluation Rubric / Model Answer null ### 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

left( xy - 5x^2sqrt1 + x^2 right) dx + (1 + x^2) dy = 0

with initial condition

y(0) = 0

. Then

y(sqrt3)

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

### 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

Q75 jee_main_2025_28_jan_evening 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 jee_main_2025_29_jan_morning 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

Q14 jee_main_2024_01_february_morning First Order Differential Equations

Let

y = y(x)

be the solution of the differential equation

fracdydx = 2x(x+y)^3 - x(x+y) - 1

, with

y(0) = 1

. Then,

left( frac1sqrt2 + yleft(frac1sqrt2right) right)^2

equals:

A. $frac44+sqrte$ $\frac{4}{4+\sqrt{e}}$
B. $frac33-sqrte$ $\frac{3}{3-\sqrt{e}}$
C. $frac21+sqrte$ $\frac{2}{1+\sqrt{e}}$
D. $frac12-sqrte$ $\frac{1}{2-\sqrt{e}}$

Solution

### Related Formula For differential equations where terms are functions of a linear expression like

(x+y)

, substitute a new variable

t = x+y

to enable variable separation. ### Core Logic Given the differential equation:

fracdydx = 2x(x+y)^3 - x(x+y) - 1

Let

x+y = t implies 1 + fracdydx = fracdtdx implies fracdydx = fracdtdx - 1

. Substituting these terms back:

fracdtdx - 1 = 2xt^3 - xt - 1

fracdtdx = 2xt^3 - xt implies fracdtdx = xt(2t^2 - 1)

### Step 1: Separating Variables and Integrating Separating variables:

fracdtt(2t^2 - 1) = x \, dx

Multiply numerator and denominator by

t

fract \, dtt^2(2t^2 - 1) = x \, dx

Let

t^2 = z implies 2t \, dt = dz implies t \, dt = fracdz2

int fracdz2z(2z-1) = int x \, dx implies int fracdzz(2z-1) = int 2x \, dx

Using partial fractions:

int left( frac22z-1 - frac1z right) dz = int 2x \, dx

ln|2z-1| - ln|z| = x^2 + C implies lnleft|frac2z-1zright| = x^2 + C

### Step 2: Apply the Boundary Condition Given

y(0) = 1 implies

x = 0

y = 1 implies t = 0 + 1 = 1 implies z = t^2 = 1

. Substituting these value constraints into our integral solution:

lnleft|frac2(1)-11right| = 0^2 + C implies ln(1) = C implies C = 0

Thus:

frac2z-1z = e^x^2 implies 2 - frac1z = e^x^2 implies frac1z = 2 - e^x^2 implies z = frac12 - e^x^2

Since

z = t^2 = (x+y)^2

, we have:

(x+y)^2 = frac12 - e^x^2

### Step 3: Evaluate at the Target Value We need to find the value of the function at

x = frac1sqrt2

left( frac1sqrt2 + yleft(frac1sqrt2right) right)^2 = frac12 - e^left(frac1sqrt2right)^2 = frac12 - e^1/2 = frac12 - sqrte

### Pattern Recognition Sees: Differential equation format

y' = f(x+y)

. Shortcut: A linear argument

(x+y)

strongly implies substituting

t=x+y

. The final expression requested matched the functional template

(x+y)^2

exactly, saving steps from extracting standalone square roots. ### 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 curve of the differential equation $(1 + x^2)dy + (y - tan^-1x)dx = 0$ $(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$ , $y(0) = 1$ $y(0) = 1$ . Then the value of $y(1)$ $y(1)$ is:

Solution & Explanation

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Let y = y(x)$y = y(x)$ be the solution curve of the differential equation (1 + x^2)dy + (y - tan^-1x)dx = 0$(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$ , y(0) = 1$y(0) = 1$ . Then the value of y(1)$y(1)$ is:

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Let $y = y(x)$ $y = y(x)$ be the solution curve of the differential equation $(1 + x^2)dy + (y - tan^-1x)dx = 0$ $(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$ , $y(0) = 1$ $y(0) = 1$ . Then the value of $y(1)$ $y(1)$ is: