If the solution of the differential equation (2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3, is alpha x+beta y+3 |2x+3y-gamma|=6…

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q66 jee_main_2025_28_jan_morning Leibniz Rule and Linear Differential Equations

Let for some function

y = f(x)

,

\int₀^x t f(t) dt = x² f(x), x > 0 and f(2) = 3.

Then

f(6)

is equal to: (1) 1 (2) 2 (3) 6 (4) 3

A. 1
B. 2
C. 6
D. 3

Solution

Related Formula

Leibniz Rule for differentiating under the integral sign:

(d)/(dx) [ \intψ(x)φ(x) f(t) dt ] = f(φ(x))φ^ (x) - f(ψ(x))ψ^ (x)

Core Logic

Differentiate both sides of the integral equation with respect to $x$ $x$:

xf(x) = x² f^ (x) + 2xf(x) -xf(x) = x² f^ (x)

Step 1: Solving the Separable Differential Equation

Separating variables:

\int (f^ (x))/(f(x)) dx = \int -(1)/(x) dx ln |f(x)| = -ln x + ln c f(x) = (c)/(x)

Step 2: Applying Boundary Constraints

Given $f(2) = 3$ $f(2) = 3$:

3 = (c)/(2) c = 6 f(x) = (6)/(x)

Evaluating for $x = 6$ $x = 6$:

f(6) = (6)/(6) = 1

Pattern Recognition

Differentiating integral statements instantly converts complex integral equations into clean, separable differential equations.

Chapter Mix

Class 12 Maths: Differential Equations

Q62 jee_main_2025_03_april_morning Linear Differential Equations

Let

g

be a differentiable function such that

\int₀xg(t)dt=x-\int₀xtg(t)dt

,

x\geq0

and let

y=y(x)

satisfy the differential equation

(dy)/(dx) - y x = 2(x+1) x g(x)

,

xin[0,(π)/(2))

. If

y(0)=0

then

y((π)/(3))

is equal to:

A. $2π3\sqrt(3)$ $\frac{2\pi}{3\sqrt{3}}$
B. $(4π)/(3)$ $\frac{4\pi}{3}$
C. $(2π)/(3)$ $\frac{2\pi}{3}$
D. $4π3\sqrt(3)$ $\frac{4\pi}{3\sqrt{3}}$

Solution

Related Formula

Leibniz Integral Rule for differentiation:

ddx(\int₀^x f(t)dt) = f(x)

Core Logic

Differentiate the given integral relation using Leibniz rule:

ddx[\int₀xg(t)dt] = ddx[x-\int₀xtg(t)dt]

g(x) = 1 - xg(x) g(x)(1+x) = 1 g(x) = (1)/(1+x)

Substitute $g(x)$ $g(x)$ into target differential equation configuration:

dydx - y x = 2(x+1) x \cdot ((1)/(1+x)) = 2 x

Step 1: Finding the Integrating Factor

This matches a linear form $dydx + P(x)y = Q(x)$ where $P(x) = - x$ $P(x) = -\tan x$.

I.F. = e\int - x dx = eln| x| = x

Write general functional solution template:

y \cdot x = \int (2 x \cdot x) dx = \int 2 dx = 2x + C

Given boundary condition $y(0) = 0 0 = 0 + C C = 0$ .

y(x) = (2x)/( x) = 2x x

Step 2: Numeric substitution

Substitute variable parameter values $x = (π)/(3)$ $x = \frac{\pi}{3}$:

y((π)/(3)) = 2((π)/(3)) ((π)/(3)) = (2π)/(3) \cdot 2 = (4π)/(3)

Pattern Recognition

Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q66 jee_main_2025_04_april_evening Linear Differential Equations

If a curve

y = y(x)

passes through the point

(1, (π)/(2))

and satisfies the differential equation

(7x⁴ y - e^x cosec y) (dx)/(dy) = x⁵

,

x \geq 1

, then at

x = 2

, the value of cosine is:

A. $2e² - e64$ $\frac{2\mathrm{e}^2 - \mathrm{e}}{64}$
B. $2e² + e64$ $\frac{2\mathrm{e}^2 + \mathrm{e}}{64}$
C. $2e² - e128$
D. $2e² + e128$

Solution

Core Logic

Let's rearrange the given differential equation by expressing $(dy)/(dx)$ $\frac{dy}{dx}$:

x⁵ (dy)/(dx) = 7x⁴ y - e^x y

Dividing both sides by $x⁵$ $x^5$:

(dy)/(dx) = (7)/(x) y - (e^x)/(x⁵) y

Multiply the entire equation by $y$ $\sin y$ to clear denominators:

y (dy)/(dx) - (7)/(x) y = -(e^x)/(x⁵)

This can be transformed into a linear form by substituting $t = - y$ $t = -\cos y$. Then $(dt)/(dx) = y (dy)/(dx)$ $\frac{dt}{dx} = \sin y \frac{dy}{dx}$.

Step 1: Solving the Linear ODE

Substituting $t$ $t$ leads to:

(dt)/(dx) + (7)/(x) t = -(e^x)/(x⁵)

This is a standard linear first-order ODE with $P(x) = (7)/(x)$ $P(x) = \frac{7}{x}$. The Integrating Factor (I.F.) is:

I.F. = e\int (7)/(x) dx = e7 ln x = x⁷

The general solution is:

t \cdot x⁷ = \int (-(e^x)/(x⁵)) \cdot x⁷ dx = -\int x² e^x dx

Step 2: Evaluating the Integration and Constant

Using integration by parts for $\int x² e^x dx$ $\int x^2 e^x dx$:

\int x² e^x dx = x² e^x - 2xe^x + 2e^x

Substituting this back:

- y \cdot x⁷ = -e^x(x² - 2x + 2) + C

y \cdot x⁷ = e^x(x² - 2x + 2) - C

Since the curve passes through $(1, (π)/(2))$ $\left(1, \frac{\pi}{2}\right)$:

((π)/(2)) \cdot (1)⁷ = e¹(1² - 2(1) + 2) - C 0 = e(1) - C C = e

Step 3: Calculating cos y at x = 2

Now substitute $x = 2$ $x = 2$ and $C = e$ $C = e$ into our equation block:

y \cdot (2⁷) = e²(2² - 2(2) + 2) - e

y \cdot 128 = e²(4 - 4 + 2) - e = 2e² - e

y = (2e² - e)/(128)

Pattern Recognition

When trigonometric terms are mixed inside an ODE containing derivative blocks like $(dx)/(dy)$ $\frac{dx}{dy}$ or $(dy)/(dx)$ $\frac{dy}{dx}$, check if clearing denominators using $y$ $\sin y$ or $y$ $\cos y$ reveals a standard substitution path for a Linear ODE.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q58 jee_main_2025_04_april_morning Area Bounded by Curves

Let

f [0, \infty) \to R

be a differentiable function such that

f(x) = 1 - 2x + \int₀^x ex - t f(t) dt

for all

x in [0, \infty)

. Then the area of the region bounded by

y = f(x)

and the coordinate axes is

A. $\sqrt(5)$ $\sqrt{5}$
B. $(1)/(2)$ $\frac{1}{2}$
C. $\sqrt(2)$ $\sqrt{2}$
D. 2

Solution

Related Formula

Leibniz Integral Rule for differentiation under integral sign:

ddx(\intφ(x)ψ(x) f(t)dt) = f(ψ(x))ψ'(x) - f(φ(x))φ'(x)

Core Logic

Rewrite equation to isolate the integral kernel:

y = 1 - 2x + e^x \int₀^x e-t f(t)dt

Differentiating with respect to $x$ $x$ using product rule and Leibniz rule:

dydx = -2 + e^x \int₀^x e-t f(t)dt + e^x \cdot (e-x f(x))

Notice that $e^x \int₀^x e-t f(t)dt = y - (1 - 2x)$ . Substitute this back:

dydx = -2 + [y - 1 + 2x] + y dydx - 2y = 2x - 3

Step 1: Solve Differential Equation

Integrating factor $I.F. = e\int -2 dx = e-2x$ .

y e-2x = \int (2x - 3)e-2xdx = (-(2x - 3))/(2)e-2x - (1)/(2)e-2x + c

From original equation, at $x=0$ $x=0$, $f(0) = 1$ $f(0) = 1$. Evaluating $c$ $c$:

1 = (3)/(2) - (1)/(2) + c c = 0

Thus, $y = -x + 1 x + y = 1$ $y = -x + 1 \implies x + y = 1$.

Step 2: Calculate Area

The boundary line is $x + y = 1$ $x + y = 1$. The area bounded by this line and coordinate axes is a right triangle with intercepts $(1,0)$ $(1,0)$ and $(0,1)$ $(0,1)$:

Area = (1)/(2) \times 1 \times 1 = (1)/(2)

Pattern Recognition

Integral equations of convolution type ( $ex-t$ $e^{x-t}$) always simplify directly into standard linear ordinary differential equations of first or second order when tracking Leibniz rules properly.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Area Under Curves

Q63 jee_main_2025_07_april_evening Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(x² + 1)y' - 2xy = (x⁴ + 2x² + 1) x

,

y(0) = 1

. Then

\int₋₃³ y(x) dx

is:

A. $24$ $24$
B. $36$ $36$
C. $30$ $30$
D. $18$ $18$

Solution

Related Formula

For a linear differential equation $(dy)/(dx) + Py = Q$ $\frac{dy}{dx} + Py = Q$, the Integrating Factor (IF) is defined as:

IF = e\int P dx

Core Logic

Divide the full differential equation by $(x²+1)$ $(x^2+1)$:

(dy)/(dx) - ((2x)/(x²+1))y = ((x²+1)² x)/(x²+1) = (x²+1) x

This is a standard Linear Differential Equation with:

P = -(2x)/(x²+1), Q = (x²+1) x

IF = e\int -(2x)/(x²+1) dx = e-ln(x²+1) = (1)/(x²+1)

Step 1: Solve for General Solution

The solution format is $y \cdot IF = \int Q \cdot IF dx$ :

y \cdot (1)/(x²+1) = \int (x²+1) x \cdot (1)/(x²+1) dx

(y)/(x²+1) = x + c

Using the boundary condition $y(0) = 1$ $y(0) = 1$:

(1)/(0+1) = (0) + c c = 1

y = (x²+1)(sin x + 1)

Step 2: Definite Integration Evaluation

We need to evaluate $\int₋₃³ y dx$ $\int_{-3}^{3} y \, dx$:

\int₋₃³ (x²+1)(sin x + 1) dx = \int₋₃³ (x² x + x² + x + 1) dx

By symmetry of odd/even functions over symmetric intervals $[-a, a]$ $[-a, a]$: $\int₋₃³ x² x dx = 0$ $\int_{-3}^{3} x^2\sin x \, dx = 0$ (since it is an odd function) $\int₋₃³ x dx = 0$ $\int_{-3}^{3} \sin x \, dx = 0$ (since it is an odd function)

Thus, we are left with the even components:

\int₋₃³ (x² + 1) dx = 2 \int₀³ (x² + 1) dx = 2 [ (x³)/(3) + x ]₀³ = 2(9 + 3) = 24

Pattern Recognition

Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus

If the solution of the differential equation (2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3$(2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3$, is α x+β y+3 ₑ|2x+3y-γ|=6$\alpha x+\beta y+3 \log_{e}|2x+3y-\gamma|=6$, then α+2β+3γ$\alpha+2\beta+3\gamma$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Translating and Simplifying

Step 2: Variable Separation Integration

Step 3: Restoring Original Variables

Step 4: Finding the Constant of Integration

Step 5: Comparing and Final Evaluation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Solving the Separable Differential Equation

Step 2: Applying Boundary Constraints

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Integrating Factor

Step 2: Numeric substitution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Solving the Linear ODE

Step 2: Evaluating the Integration and Constant

Step 3: Calculating cos y at x = 2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve Differential Equation

Step 2: Calculate Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for General Solution

Step 2: Definite Integration Evaluation

Pattern Recognition

Chapter Mix

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If the solution of the differential equation $(2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3$ $(2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3$, is $α x+β y+3 ₑ|2x+3y-γ|=6$ , then $α+2β+3γ$ $\alpha+2\beta+3\gamma$ is equal to: