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 6

Q12 jee_main_2024_29_jan_morning Linear Differential Equations

A function

y=f(x)

satisfies

f(x)sin 2x+sin x-(1+cos^2 x)f'(x)=0

with condition

f(0)=0

. Then

f(fracpi2)

is equal to

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

Solution

### Related Formula

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

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

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

### Core Logic Given the differential equation:

ysin 2x + sin x - (1+cos^2 x)fracdydx = 0

Rearrange it into the standard linear differential equation format:

(1+cos^2 x)fracdydx - ysin 2x = sin x

fracdydx - left(fracsin 2x1+cos^2 xright)y = fracsin x1+cos^2 x

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

P(x) = -fracsin 2x1+cos^2 x

I.F. = e^int -fracsin 2x1+cos^2 x dx

Let

1+cos^2 x = t

, then

dt = -2cos xsin x \, dx = -sin 2x \, dx

I.F. = e^int frac1t dt = e^ln t = t

I.F. = 1+cos^2 x

### Step 2: Solve the Differential Equation Multiply the LDE by the integrating factor:

y cdot (1+cos^2 x) = int left(fracsin x1+cos^2 xright) cdot (1+cos^2 x) dx

y(1+cos^2 x) = int sin x dx

y(1+cos^2 x) = -cos x + C

Apply the initial condition

f(0) = 0

(which means

y=0

when

x=0

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

The particular solution is:

y(1+cos^2 x) = 1 - cos x

### Step 3: Evaluate at x = pi/2 To find

f(fracpi2)

, substitute

x = fracpi2

yleft(1+cos^2left(fracpi2right)right) = 1 - cosleft(fracpi2right)

y(1+0) = 1 - 0

y = 1

### Pattern Recognition Whenever

y

and

y'

appear linearly separated by complex trigonometric polynomials, format aggressively into standard LDE form. The derivative of

1+cos^2x

is exactly

-sin 2x

, making the logarithmic integration seamless. ### Evaluation Rubric / Model Answer null ### 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+y^2)(1+log_e x)dx+x dy=0

x gt 0

passes through the point

(1, 1)

and

y(e)=fracalpha-tan(frac32)beta+tan(frac32)

, then

alpha+2beta

Numerical Answer. Answer: 3 to 3

Solution

### Related Formula

int frac11+y^2 dy = tan^-1y + C

int f(x) f'(x) dx = frac[f(x)]^22 + C

tan(A - B) = fractan A - tan B1 + tan A tan B

### Core Logic Rearrange the given differential equation to separate variables

x

and

y

(1+y^2)(1+log_e x)dx + x dy = 0

frac1+log_e xx dx + frac11+y^2 dy = 0

Integrate both sides:

int frac1+log_e xx dx + int frac11+y^2 dy = C

### Step 1: Execute Integration For the first integral, let

u = 1+log_e x

, so

du = frac1x dx

int u \, du = fracu^22 = frac(1+log_e x)^22

For the second integral:

int frac11+y^2 dy = tan^-1y

Putting them together:

frac(1+log_e x)^22 + tan^-1y = C

### Step 2: Apply Boundary Conditions The curve passes through

(1, 1)

. Substitute

x=1, y=1

frac(1+log_e 1)^22 + tan^-1(1) = C

Since

log_e 1 = 0

and

tan^-1(1) = fracpi4

frac1^22 + fracpi4 = C Rightarrow C = frac12 + fracpi4

Wait, the solution rearranges differently:

int (frac1x + fracln xx) dx = ln x + frac(ln x)^22

. Let's use this to stay fully synced with the PDF's algebraic step.

ln x + frac(ln x)^22 + tan^-1 y = C

(1,1)

ln(1) + frac(ln 1)^22 + tan^-1(1) = C Rightarrow 0 + 0 + fracpi4 = C Rightarrow C = fracpi4

The general equation simplifies nicely to:

ln x + frac(ln x)^22 + tan^-1 y = fracpi4

### Step 3: Evaluate at x = e Substitute

x=e

into the equation to find

y(e)

ln e + frac(ln e)^22 + tan^-1 y = fracpi4

1 + frac1^22 + tan^-1 y = fracpi4

frac32 + tan^-1 y = fracpi4

tan^-1 y = fracpi4 - frac32

y = tanleft(fracpi4 - frac32right)

Using the compound angle formula

tan(A-B)

y = fractan(pi/4) - tan(3/2)1 + tan(pi/4)tan(3/2) = frac1 - tan(3/2)1 + tan(3/2)

Comparing this format to

y(e) = fracalpha-tan(3/2)beta+tan(3/2)

gives:

alpha = 1, quad beta = 1

Calculate

alpha + 2beta

1 + 2(1) = 3

### Pattern Recognition For expressions involving separated

x

and

y

logarithmic terms, always split

int frac1+ln xxdx

into

int frac1x + int fracln xx

to avoid carrying a non-zero shift

frac12

into the constant

C

. This allows direct identity matching for tangent sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

Q15 jee_main_2024_30_january_evening Functional Equations

Let

f : mathbbR - \0\ to mathbbR

be a function satisfying

fleft(fracxyright) = fracf(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

textChain rule for partial differentiation: fracpartialpartial x fleft(fracxyright) = f'left(fracxyright) cdot frac1y

### Core Logic Given functional equation:

fleft(fracxyright) = fracf(x)f(y)

First, put

x=1, y=1

f(1) = fracf(1)f(1) Rightarrow f(1) = 1

(since

f(y) neq 0

implies

f(1) neq 0

). ### Step 1: Partial Differentiation Partially differentiate the given equation with respect to

x

, keeping

y

constant:

f'left(fracxyright) cdot frac1y = frac1f(y) cdot f'(x)

Let

x to y

so that

fracxy to 1

. Wait, let's substitute

y = x

directly:

f'(1) cdot frac1x = frac1f(x) cdot f'(x)

### Step 2: Forming the Final Equation We are given

f'(1) = 2024

. Substituting this value:

2024 cdot frac1x = fracf'(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)

inherently describe power functions (

f(x) = x^k

). Partial differentiation with respect to one variable quickly resolves the derivative form without limits. ### Evaluation Rubric / Model Answer null ### 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

frac-y^22Y'(x) + 1

Y'(x) neq 0

. If

Y(1) = 1

, then

12Y(2)

equals

Numerical Answer. Answer: 20 to 20

Solution

### Related Formula

textEquation of tangent at (x, y): Y - y = fracdydx(X - x)

textArea of Right Triangle: frac12 times textbase times textheight

### Core Logic The line

Y - y = y'(X - x)

intersects the axes to form a triangle. Let's find the intercepts: X-intercept (set

Y=0

-y = y'(X - x) Rightarrow X = x - fracyy'

Y-intercept (set

X=0

Y - y = y'(-x) Rightarrow Y = y - xy'

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

textArea = frac12 left( x - fracyy' right) (y - xy') = frac-y^22y' + 1

Linear Differential Equation diagram for Q21 - JEE Main 2024 Evening

### Step 1: Setting up the Differential Equation Expand the Area equation:

frac12 left[ xy - x^2y' - fracy^2y' + xy right] = frac-y^22y' + 1

Multiply by 2:

2xy - x^2y' - fracy^2y' = -fracy^2y' + 2

Cancel

-fracy^2y'

from both sides:

2xy - x^2y' = 2

### Step 2: Solving the Linear Differential Equation Rearrange into standard LDE form:

x^2 fracdydx - 2xy = -2

fracdydx - frac2xy = -frac2x^2

Find the Integrating Factor (IF):

textI.F. = e^int -frac2x dx = e^-2ln x = x^-2 = frac1x^2

Multiply the LDE by the IF:

y cdot frac1x^2 = int left( -frac2x^2 right) frac1x^2 dx + C

y x^-2 = -2 int x^-4 dx + C

y x^-2 = -2 left( fracx^-3-3 right) + C

fracyx^2 = frac23x^3 + C

y = frac23x + Cx^2

### Step 3: Applying Initial Conditions Given

Y(1) = 1

(i.e., when

x=1, y=1

1 = frac23(1) + C(1)^2 Rightarrow 1 = frac23 + C Rightarrow C = frac13

Thus, the curve is:

y = frac23x + fracx^23

We need to find

12Y(2)

Y(2) = frac23(2) + frac2^23 = frac13 + frac43 = frac53

12Y(2) = 12 left( frac53 right) = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Application of Derivatives

Q9 jee_main_2024_30_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

sec x \, dy + \2(1 - x) tan x + x(2 - x)\ \, dx = 0

such that

y(0) = 2

. Then

y(2)

is equal to:

A. $2$ $2$
B. $2\1 - sin (2)\$ $2\{1 - \sin (2)\}$
C. $2\sin (2) + 1\$ $2\{\sin (2) + 1\}$
D. $1$ $1$

Solution

### Related Formula

d(u cdot v) = u \, dv + v \, du

### Core Logic Given the differential equation:

sec x \, dy + \2(1 - x) tan x + (2x - x^2)\ \, dx = 0

Divide entirely by

sec x

(which is

1/cos x

dy + \2(1 - x) sin x + (2x - x^2) cos x\ \, dx = 0

Rearranging it as an exact differential form:

fracdydx = 2(x - 1) sin x + (x^2 - 2x) cos x

### Step 1: Integration using By-Parts Now integrate both sides:

y(x) = int 2(x - 1) sin x \, dx + int (x^2 - 2x) cos x \, dx

Apply integration by parts on the second integral, where

u = x^2 - 2x

and

dv = cos x \, dx

int (x^2 - 2x) cos x \, dx = (x^2 - 2x)(sin x) - int (2x - 2) sin x \, dx

Substituting this back into

y(x)

y(x) = int 2(x - 1) sin x \, dx + left[ (x^2 - 2x) sin x - int 2(x - 1) sin x \, dx right] + lambda

The integrals cancel out perfectly:

y(x) = (x^2 - 2x) sin x + lambda

### Step 2: Finding Constant and Final Value Use the initial condition

y(0) = 2

y(0) = 0 + lambda Rightarrow lambda = 2

Thus, the solution is:

y(x) = (x^2 - 2x) sin x + 2

To find

y(2)

y(2) = (2^2 - 2(2)) sin 2 + 2 = (4 - 4) sin 2 + 2 = 2

### Pattern Recognition When an integrand is a mix of polynomial and trigonometric terms, grouping them to form an exact differential or observing that integration by parts on one term perfectly cancels the other term is a classic structural trick. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Integrals

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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: