Let f: R R be a thrice differentiable odd function satisfying f'(x) geq 0, f'(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f(…

Q27 jee_main_2024_30_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(1 - x^2) dy = left[ xy + (x^3 + 2)sqrt3(1 - x^2) right] dx, quad -1 < x < 1, y(0) = 0

. If

yleft(frac12right) = fracmn

,

m

and

n

are co-prime numbers, then

m + n

is equal to

Numerical Answer. Answer: 97 to 97

Solution

### Related Formula For a linear differential equation

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

:

IF = e^int P(x) dx

Solution:

y cdot IF = int Q(x) cdot IF \, dx + C

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

(1 - x^2) fracdydx = xy + (x^3 + 2)sqrt3(1 - x^2)

fracdydx - fracx1 - x^2 y = frac(x^3 + 2)sqrt3(1 - x^2)1 - x^2

fracdydx - fracx1 - x^2 y = fracsqrt3(x^3 + 2)sqrt1 - x^2

Identify

P(x) = -fracx1 - x^2

. ### Step 1: Finding Integrating Factor (IF)

IF = e^int -fracx1 - x^2 dx

Let

1 - x^2 = t Rightarrow -2x dx = dt Rightarrow -x dx = fracdt2

.

IF = e^frac12 int frac1t dt = e^frac12 ln t = e^ln sqrtt = sqrt1 - x^2

### Step 2: General Solution The solution is given by:

y sqrt1 - x^2 = int left( fracsqrt3(x^3 + 2)sqrt1 - x^2 right) sqrt1 - x^2 \, dx + C

y sqrt1 - x^2 = sqrt3 int (x^3 + 2) \, dx + C

y sqrt1 - x^2 = sqrt3 left( fracx^44 + 2x right) + C

### Step 3: Finding Constant C Using

y(0) = 0

:

0 cdot 1 = sqrt3(0 + 0) + C Rightarrow C = 0

So,

y(x) = fracsqrt3sqrt1 - x^2 left( fracx^44 + 2x right)

. ### Step 4: Evaluating required point Substitute

x = 1/2

:

yleft(frac12right) = fracsqrt3sqrt1 - 1/4 left( frac1/164 + 2left(frac12right) right)

Wait,

frac(1/2)^44 = frac1/164 = frac164

. Let's recheck the expression:

yleft(frac12right) = fracsqrt3sqrt3/4 left( frac164 + 1 right)

= fracsqrt3sqrt3/2 left( frac6564 right) = 2 times frac6564 = frac6532

Here,

m = 65

and

n = 32

. They are co-prime. Thus,

m + n = 65 + 32 = 97

. ### Pattern Recognition Whenever roots matching the integration denominator appear on the RHS of a linear differential setup, it is a high-confidence signal that the integrating factor cleanly annihilates the fractional root component during the solution stage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

Q8 jee_main_2024_31_jan_evening Newton's Law of Cooling

The temperature

T(t)

of a body at time

t = 0

is

160^circ mathrmF

and it decreases continuously as per the differential equation

fracmathrmdTmathrmdt = -mathrmK(T - 80)

, where

K

is positive constant. If

T(15) = 120^circmathrmF

, then

T(45)

is equal to

A. $85^circmathrmF$ $85^{\circ}\mathrm{F}$
B. $95^circmathrmF$ $95^{\circ}\mathrm{F}$
C. $90^mathrmomathrmF$ $90^{\mathrm{o}}\mathrm{F}$
D. $80^circmathrmF$ $80^{\circ}\mathrm{F}$

Solution

### Related Formula

int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C

### Core Logic Given

fracdTdt = -K(T-80)

. Integrating from

t=0

to

t

:

int_160^T fracdTT-80 = -K int_0^t dt

[ln|T-80|]_160^T = -Kt

lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt

Given

T(15) = 120

:

120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12

To find

T(45)

:

T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3

= 80 + 80left(frac12right)^3

= 80 + 80left(frac18right) = 90

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

Q29 jee_main_2024_31_jan_evening Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

sec^2 x \, dx + left( e^2y tan^2 x + tan x right) dy = 0

,

0 < x < fracpi2

,

yleft(fracpi4right) = 0

. If

yleft(fracpi6right) = alpha

. Then

e^8alpha

is equal to

Numerical Answer. Answer: 9 to 9

Solution

### Related Formula

textIntegrating factor for fracdudy + P(y)u = Q(y) text is I.F. = e^int P(y)dy

### Core Logic Given DE:

sec^2 x fracdxdy + e^2y tan^2 x + tan x = 0

Substitute

t = tan x implies fracdtdy = sec^2 x fracdxdy

. The DE becomes:

fracdtdy + t = -t^2 e^2y

This is a Bernoulli equation in

t

. Divide by

t^2

:

frac1t^2fracdtdy + frac1t = -e^2y

Substitute

u = frac1t implies fracdudy = -frac1t^2fracdtdy

.

-fracdudy + u = -e^2y implies fracdudy - u = e^2y

This is a linear DE in

u

with respect to

y

.

P(y) = -1, Q(y) = e^2y

. Integrating Factor:

I.F. = e^int -1 dy = e^-y

. Solution:

u e^-y = int e^2y e^-y dy = int e^y dy = e^y + C

Substitute

u = frac1tan x

:

frace^-ytan x = e^y + C

Use given condition

y(pi/4) = 0

:

frace^0tan(pi/4) = e^0 + C implies 1 = 1 + C implies C = 0

Therefore,

frace^-ytan x = e^y implies tan x = e^-2y

. Evaluate at

x = pi/6

,

y = alpha

:

tan(pi/6) = e^-2alpha implies frac1sqrt3 = e^-2alpha

e^2alpha = sqrt3 implies (e^2alpha)^4 = (sqrt3)^4 = 9

Thus,

e^8alpha = 9

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

Q9 jee_main_2024_31_jan_morning Homogeneous Differential Equations

The solution curve of the differential equation

yfracdxdy = x(log_e x - log_e y + 1), x > 0, y > 0

passing through the point

(e, 1)

is

A. $|log_e fracyx| = x$ $$|\log_e \frac{y}{x}| = x$$
B. $|log_e fracyx| = y^2$ $$|\log_e \frac{y}{x}| = y^2$$
C. $|log_e fracxy| = y$ $$|\log_e \frac{x}{y}| = y$$
D. $2|log_e fracxy| = y + 1$ $$2|\log_e \frac{x}{y}| = y + 1$$

Solution

### Core Logic Given DE:

fracdxdy = fracxy left(lnleft(fracxyright) + 1right)

Let

fracxy = t implies x = ty

. Differentiating w.r.t

y

:

fracdxdy = t + yfracdtdy

### Step 1: Substitution and Integration

t + yfracdtdy = t(ln(t) + 1) = tln t + t

yfracdtdy = tln t implies fracdttln t = fracdyy

Integrate both sides. Let

ln t = p implies frac1t dt = dp

.

int fracdpp = int fracdyy

ln|p| = ln y + C implies ln|ln t| = ln y + C

lnleft|lnleft(fracxyright)right| = ln y + C

### Step 2: Applying Boundary Conditions Given curve passes through

(e, 1)

:

lnleft|lnleft(frace1right)right| = ln(1) + C implies C = 0

lnleft|lnleft(fracxyright)right| = ln y

left|lnleft(fracxyright)right| = e^ln y = y

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

Q11 jee_main_2024_31_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

fracdydx = frac(tan x) + ysin x(sec x - sin x tan x), x in left(0, fracpi2right)

satisfying the condition

yleft(fracpi4right) = 2

. Then,

yleft(fracpi3right)

is

A. $sqrt3(2 + log_esqrt3)$ $$\sqrt{3}(2 + \log_e\sqrt{3})$$
B. $fracsqrt32(2 + log_e 3)$ $$\frac{\sqrt{3}}{2}(2 + \log_e 3)$$
C. $sqrt3(1 + 2log_e 3)$ $$\sqrt{3}(1 + 2\log_e 3)$$
D. $sqrt3(2 + log_e 3)$ $$\sqrt{3}(2 + \log_e 3)$$

Solution

### Core Logic

fracdydx = fracfracsin xcos x + ysin x left(frac1cos x - fracsin^2 xcos xright) = fracsin x + ycos xsin x (1 - sin^2 x)

fracdydx = fracsin x + ycos xsin x cos^2 x = sec^2 x + frac2ysin 2x

fracdydx - 2csc(2x)y = sec^2 x

### Step 1: Integrating Factor This is an LDE of form

fracdydx + Py = Q

.

I.F. = e^int -2csc(2x) dx

Let

2x = t implies 2dx = dt

.

I.F. = e^-int csc t dt = e^-ln|tan(t/2)| = e^-ln|tan x| = frac1|tan x|

### Step 2: Solution of LDE

y(I.F.) = int Q(I.F.) dx + C

yfrac1tan x = int sec^2 x frac1tan x dx + C

Let

tan x = t implies sec^2 x dx = dt

.

yfrac1tan x = int fracdtt + C = ln|tan x| + C

y = tan x(ln|tan x| + C)

### Step 3: Boundary Value Given

y(pi/4) = 2

:

2 = 1(ln 1 + C) implies C = 2

Thus,

y = tan x (ln|tan x| + 2)

. At

x = pi/3

:

y(pi/3) = sqrt3(lnsqrt3 + 2)

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

Let $f: mathbbR to mathbbR$ $f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) geq 0$ $f'(x) \geq 0$, $f'(x) = f(x)$ $f'(x) = f(x)$, $f(0) = 0$ $f(0) = 0$, $f'(0) = 3$ $f'(0) = 3$. Then $9f(log_e 3)$ $9f(\log_e 3)$ is equal to ________.

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 7

Solution

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_morning

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Let f: mathbbR to mathbbR$f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying f'(x) geq 0$f'(x) \geq 0$, f'(x) = f(x)$f'(x) = f(x)$, f(0) = 0$f(0) = 0$, f'(0) = 3$f'(0) = 3$. Then 9f(log_e 3)$9f(\log_e 3)$ is equal to ________.

Solution & Explanation

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 7

Solution

Solution

Solution

Solution

Solution

More Differential Equations Questions — jee_main_2025_02_april_morning

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Welcome back

Let $f: mathbbR to mathbbR$ $f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) geq 0$ $f'(x) \geq 0$, $f'(x) = f(x)$ $f'(x) = f(x)$, $f(0) = 0$ $f(0) = 0$, $f'(0) = 3$ $f'(0) = 3$. Then $9f(log_e 3)$ $9f(\log_e 3)$ is equal to ________.