Let f be a twice differentiable non-negative function such that (f(x))2 = 25 + 0x ( (f(t))2 + (f'(t))2 ) dt. Then the…

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Q9 jee_main_2024_30_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

x dy + 2(1 - x) x + x(2 - x) dx = 0

such that

y(0) = 2

. Then

y(2)

is equal to:

A. $2$ $2$
B. $21 - (2)$ $2\{1 - \sin (2)\}$
C. $2 (2) + 1$ $2\{\sin (2) + 1\}$
D. $1$ $1$

Solution

Related Formula

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

Core Logic

Given the differential equation:

x dy + 2(1 - x) x + (2x - x²) dx = 0

Divide entirely by $x$ $\sec x$ (which is $1/ x$ $1/\cos x$):

dy + 2(1 - x) x + (2x - x²) x dx = 0

Rearranging it as an exact differential form:

(dy)/(dx) = 2(x - 1) x + (x² - 2x) x

Step 1: Integration using By-Parts

Now integrate both sides:

y(x) = \int 2(x - 1) x dx + \int (x² - 2x) x dx

Apply integration by parts on the second integral, where $u = x² - 2x$ $u = x^2 - 2x$ and $dv = x dx$ $dv = \cos x \, dx$:

\int (x² - 2x) x dx = (x² - 2x)( x) - \int (2x - 2) x dx

Substituting this back into $y(x)$ $y(x)$:

y(x) = \int 2(x - 1) x dx + [ (x² - 2x) x - \int 2(x - 1) x dx ] + λ

The integrals cancel out perfectly:

y(x) = (x² - 2x) x + λ

Step 2: Finding Constant and Final Value

Use the initial condition $y(0) = 2$ $y(0) = 2$:

y(0) = 0 + λ \Rightarrow λ = 2

Thus, the solution is:

y(x) = (x² - 2x) x + 2

To find $y(2)$ $y(2)$:

y(2) = (2² - 2(2)) 2 + 2 = (4 - 4) 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.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Integrals

Q27 jee_main_2024_30_jan_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(1 - x²) dy = [ xy + (x³ + 2)\sqrt(3(1 - x²)) ] dx, -1 < x < 1, y(0) = 0

. If

y((1)/(2)) = (m)/(n)

,

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 $(dy)/(dx) + P(x)y = Q(x)$ $\frac{dy}{dx} + P(x)y = Q(x)$:

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 - x²) (dy)/(dx) = xy + (x³ + 2)\sqrt(3(1 - x²))

(dy)/(dx) - (x)/(1 - x²) y = (x³ + 2)\sqrt(3(1 - x²))1 - x²

(dy)/(dx) - (x)/(1 - x²) y = \sqrt(3)(x³ + 2)\sqrt(1 - x²)

Identify $P(x) = -(x)/(1 - x²)$ $P(x) = -\frac{x}{1 - x^2}$.

Step 1: Finding Integrating Factor (IF)

IF = e\int -(x)/(1 - x²) dx

Let $1 - x² = t \Rightarrow -2x dx = dt \Rightarrow -x dx = (dt)/(2)$ .

IF = e(1)/(2) \int (1)/(t) dt = e(1)/(2) ln t = eln \sqrt(t) = \sqrt(1 - x²)

Step 2: General Solution

The solution is given by:

y \sqrt(1 - x²) = \int ( \sqrt(3)(x³ + 2)\sqrt(1 - x²) ) \sqrt(1 - x²) dx + C

y \sqrt(1 - x²) = \sqrt(3) \int (x³ + 2) dx + C

y \sqrt(1 - x²) = \sqrt(3) ( (x⁴)/(4) + 2x ) + C

Step 3: Finding Constant C

Using $y(0) = 0$ $y(0) = 0$:

0 \cdot 1 = \sqrt(3)(0 + 0) + C \Rightarrow C = 0

So, $y(x) = \sqrt(3)\sqrt(1 - x²) ( (x⁴)/(4) + 2x )$ .

Step 4: Evaluating required point

Substitute $x = 1/2$ $x = 1/2$:

y((1)/(2)) = \sqrt(3)\sqrt(1 - 1/4) ( (1/16)/(4) + 2((1)/(2)) )

Wait, $((1/2)⁴)/(4) = (1/16)/(4) = (1)/(64)$ . Let's recheck the expression:

y((1)/(2)) = \sqrt(3)\sqrt(3/4) ( (1)/(64) + 1 )

= \sqrt(3)\sqrt(3)/2 ( (65)/(64) ) = 2 \times (65)/(64) = (65)/(32)

Here, $m = 65$ $m = 65$ and $n = 32$ $n = 32$. They are co-prime. Thus, $m + n = 65 + 32 = 97$ $m + n = 65 + 32 = 97$.

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.

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° F

and it decreases continuously as per the differential equation

dTdt = -K(T - 80)

, where

K

is positive constant. If

T(15) = 120°F

, then

T(45)

is equal to

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

Solution

Related Formula

\int (dT)/(T-Tₛ) = -K \int dt ln|T-Tₛ| = -Kt + C

Core Logic

Given $(dT)/(dt) = -K(T-80)$ $\frac{dT}{dt} = -K(T-80)$. Integrating from $t=0$ $t=0$ to $t$ $t$:

\int₁₆₀T (dT)/(T-80) = -K \int₀^t dt

[ln|T-80|]₁₆₀^T = -Kt

ln((T-80)/(80)) = -Kt T(t) = 80 + 80e-Kt

Given $T(15) = 120$ $T(15) = 120$:

120 = 80 + 80e-15K 40 = 80e-15K e-15K = (1)/(2)

To find $T(45)$ $T(45)$:

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

= 80 + 80((1)/(2))³

= 80 + 80((1)/(8)) = 90

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

² x dx + ( e2y ² x + x ) dy = 0

,

0 < x < (π)/(2)

,

y((π)/(4)) = 0

. If

y((π)/(6)) = α

. Then

e8α

is equal to

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

Integrating factor for (du)/(dy) + P(y)u = Q(y) is I.F. = e\int P(y)dy

Core Logic

Given DE: $² x (dx)/(dy) + e2y ² x + x = 0$ Substitute $t = x (dt)/(dy) = ² x (dx)/(dy)$ . The DE becomes:

(dt)/(dy) + t = -t² e2y

This is a Bernoulli equation in $t$ $t$. Divide by $t²$ $t^2$:

(1)/(t²)(dt)/(dy) + (1)/(t) = -e2y

Substitute $u = (1)/(t) (du)/(dy) = -(1)/(t²)(dt)/(dy)$ .

-(du)/(dy) + u = -e2y (du)/(dy) - u = e2y

This is a linear DE in $u$ $u$ with respect to $y$ $y$. $P(y) = -1, Q(y) = e2y$ $P(y) = -1, Q(y) = e^{2y}$. Integrating Factor: $I.F. = e\int -1 dy = e-y$ $I.F. = e^{\int -1 dy} = e^{-y}$. Solution:

u e-y = \int e2y e-y dy = \int e^y dy = e^y + C

Substitute $u = (1)/( x)$ $u = \frac{1}{\tan x}$:

e-y x = e^y + C

Use given condition $y(π/4) = 0$ $y(\pi/4) = 0$:

(e⁰)/( (π/4)) = e⁰ + C 1 = 1 + C C = 0

Therefore, $e-y x = e^y x = e-2y$ . Evaluate at $x = π/6$ $x = \pi/6$, $y = α$ $y = \alpha$:

(π/6) = e-2α 1\sqrt(3) = e-2α

e2α = \sqrt(3) (e2α)⁴ = (\sqrt(3))⁴ = 9

Thus, $e8α = 9$ $e^{8\alpha} = 9$.

Chapter Mix

Class 12 Maths: Differential Equations

Q9 jee_main_2024_31_jan_morning Homogeneous Differential Equations

The solution curve of the differential equation

y(dx)/(dy) = x( ₑ x - ₑ y + 1), x > 0, y > 0

passing through the point

(e, 1)

is

A. $| ₑ (y)/(x)| = x$ $$|\log_e \frac{y}{x}| = x$$
B. $| ₑ (y)/(x)| = y²$ $$|\log_e \frac{y}{x}| = y^2$$
C. $| ₑ (x)/(y)| = y$ $$|\log_e \frac{x}{y}| = y$$
D. $2| ₑ (x)/(y)| = y + 1$ $$2|\log_e \frac{x}{y}| = y + 1$$

Solution

Core Logic

Given DE: $(dx)/(dy) = (x)/(y) (ln((x)/(y)) + 1)$ Let $(x)/(y) = t x = ty$ $\frac{x}{y} = t \implies x = ty$. Differentiating w.r.t $y$ $y$:

(dx)/(dy) = t + y(dt)/(dy)

Step 1: Substitution and Integration

t + y(dt)/(dy) = t(ln(t) + 1) = tln t + t

y(dt)/(dy) = tln t (dt)/(tln t) = (dy)/(y)

Integrate both sides. Let $ln t = p (1)/(t) dt = dp$ .

\int (dp)/(p) = \int (dy)/(y)

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

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

Step 2: Applying Boundary Conditions

Given curve passes through $(e, 1)$ $(e, 1)$:

ln|ln((e)/(1))| = ln(1) + C C = 0

ln|ln((x)/(y))| = ln y

|ln((x)/(y))| = eln y = y

Chapter Mix

Class 12 Maths: Differential Equations

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve the Differential Equation

Step 2: Calculate Sequence Values

Step 3: Calculate the Mean

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Integration using By-Parts

Step 2: Finding Constant and Final Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Integrating Factor (IF)

Step 2: General Solution

Step 3: Finding Constant C

Step 4: Evaluating required point

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Substitution and Integration

Step 2: Applying Boundary Conditions

Chapter Mix

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