If the solution curve y=y(x) of the differential equation (1+y2)(1+ x)dx+x dy=0, x 0 passes through the point (1, 1)…

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q6 jee_main_2026_22_january_evening Variable Separable Form

If

y = y(x)

satisfies the differential equation

16 ( x + 9\sqrt(x)) (4 + 9 + \sqrt(x)) y dy = (1 + 2 y) dx

,

x > 0

and

y(256) = (π)/(2)

,

y(49) = α

, then

2 α

is equal to:

A. $2\sqrt(2) - 1$ $2\sqrt{2} - 1$
B. $2(\sqrt(2) - 1)$ $2(\sqrt{2} - 1)$
C. $3(\sqrt(2) - 1)$ $3(\sqrt{2} - 1)$
D. $\sqrt(2) - 1$ $\sqrt{2} - 1$

Solution

\int ( y)/(1 + 2 y) dy = \int dx16 x + 9\sqrt(x) (4 + 9 + \sqrt(x))

Core Logic

Let $t = 4 + 9 + \sqrt(x)$ $t = 4 + \sqrt{9 + \sqrt{x}}$. Then:

dt = 12 9+\sqrt(x) \cdot 12\sqrt(x) dx = dx4 x(9+\sqrt(x)) = dx4 x + 9\sqrt(x)

Substitute into differential equation:

(1)/(2) ln |1 + 2 y| = \int (4 dt)/(16 t) = (1)/(4) ln t + C

(1)/(2) ln (1 + 2 y) = (1)/(4) ln (4 + 9 + \sqrt(x)) + C

Step 1: Determine Constant C

Using initial condition $y(256) = (π)/(2)$ $y(256) = \frac{\pi}{2}$:

(1)/(2) ln(3) = (1)/(4) ln(4 + \sqrt(9 + 16)) + C = (1)/(4) ln(9) + C = (1)/(2) ln 3 + C C = 0

Step 2: Calculate for x = 49

For $x = 49$ $x = 49$, $y = α$ $y = \alpha$:

(1)/(2) ln (1 + 2 α) = (1)/(4) ln(4 + \sqrt(9 + 7)) = (1)/(4) ln 8

ln(1 + 2 α) = (1)/(2) ln 8 = ln(2\sqrt(2)) 1 + 2 α = 2\sqrt(2) 2 α = 2\sqrt(2) - 1

Pattern Recognition

Identify substitution $t = 4 + 9 + \sqrt(x)$ $t = 4 + \sqrt{9 + \sqrt{x}}$ to simplify complex radical integral expression.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Indefinite Integration

Q7 jee_main_2026_23_january_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

x⁴ dy + (4x³ y + 2 x)dx = 0

,

x > 0

,

y((π)/(2)) = 0

. Then

π⁴ y((π)/(3))

is equal to:

A. $81$ $81$
B. $92$ $92$
C. $64$ $64$
D. $72$ $72$

Solution

Related Formula

d(uv) = u dv + v du

Core Logic

The differential equation can be grouped using the exact differential logic.

x⁴ dy + (4x³ y + 2 x)dx = 0

(x⁴ dy + 4x³ y dx) = -2 x dx

Notice that $x⁴ dy + 4x³ y dx = d(x⁴ y)$ $x^4 dy + 4x^3 y dx = d(x^4 y)$. So, $\int d(x⁴ y) = \int -2 x dx$ $\int d(x^4 y) = \int -2\sin x dx$.

Step 1: Integrate both sides

Integrating gives:

x⁴ y = 2 x + C

Since $y(x) = f(x)$ $y(x) = f(x)$, we have $x⁴ f(x) = 2 x + C$ $x^4 f(x) = 2\cos x + C$.

Step 2: Find Integration Constant

Use the condition $f((π)/(2)) = 0$ $f\left(\frac{\pi}{2}\right) = 0$:

((π)/(2))⁴ (0) = 2 ((π)/(2)) + C \Rightarrow C = 0

So, the solution is: $x⁴ y = 2 x$ $x^4 y = 2\cos x$

Step 3: Final Calculation

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

((π)/(3))⁴ y((π)/(3)) = 2 ((π)/(3))

(π⁴)/(81) y((π)/(3)) = 2((1)/(2)) = 1

π⁴ y((π)/(3)) = 81

Pattern Recognition

Whenever you see polynomial terms mixed with $xⁿ dy + n xⁿ⁻¹ y dx$ $x^n dy + n x^{n-1} y dx$, always condense it into the exact differential $d(xⁿ y)$ $d(x^n y)$ rather than using the standard integrating factor method.

Chapter Mix

Class 12 Maths: Differential Equations

Q23 jee_main_2026_23_january_morning Solution of Differential Equations

Let

f

be a twice differentiable non-negative function such that

(f(x))² = 25 + \int₀x ( (f(t))² + (f'(t))² ) dt.

Then the mean of

f( ₑ(1))

,

f( ₑ(2))

, ...,

f( ₑ(625))

is equal to _____.

Numerical Answer. Answer: 1565 to 1565

Solution

Related Formula

(d)/(dx)\int₀^x g(t)dt = g(x) (Leibniz Rule)

Core Logic

Differentiate both sides with respect to $x$ $x$ using Leibniz rule:

2f(x)f'(x) = f²(x) + (f'(x))²

Rearranging yields a perfect square:

(f'(x))² - 2f(x)f'(x) + f²(x) = 0

(f'(x) - f(x))² = 0

Thus, $f'(x) = f(x)$ $f'(x) = f(x)$.

Step 1: Solve the Differential Equation

Integrate $(f'(x))/(f(x)) = 1$ $\frac{f'(x)}{f(x)} = 1$:

ln f(x) = x + C

f(x) = ex+C = k e^x

To find $k$ $k$, plug $x = 0$ $x = 0$ into the original integral equation:

(f(0))² = 25 + \int₀⁰ dt = 25

Since $f$ $f$ is non-negative, $f(0) = 5$ $f(0) = 5$. Thus, $k = 5$ $k = 5$ and $f(x) = 5e^x$ $f(x) = 5e^x$.

Step 2: Calculate Sequence Values

Evaluate $f(ln n)$ $f(\ln n)$ for $n=1, 2, \dots, 625$ $n=1, 2, \ldots, 625$:

f(ln n) = 5eln n = 5n

Step 3: Calculate the Mean

Mean of $f(ln 1), f(ln 2), , f(ln 625)$ $f(\ln 1), f(\ln 2), \dots, f(\ln 625)$:

Mean = Σn=1⁶²⁵ 5n625 = (5 (1 + 2 + + 625))/(625)

= (5)/(625) \times (625 \times 626)/(2) = (5 \times 626)/(2) = 5 \times 313 = 1565

Pattern Recognition

An integral equation with $x$ $x$ as an upper limit immediately demands Leibniz differentiation. If it results in a perfect square differential relation, an exponential solution map natively emerges.

Chapter Mix

Class 12 Maths: Differential Equations

Q21 jee_main_2026_23_january_evening Linear Differential Equations

If the solution curve

y = f(x)

of the differential equation

(x² - 4)y' - 2xy + 2x(4 - x²)² = 0, x > 2

, passes through the point

(3, 15)

, then the local maximum value of

f

is

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

For a linear differential equation $(dy)/(dx) + P(x)y = Q(x)$ $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is $e\int P(x) dx$ $e^{\int P(x) dx}$.

Core Logic

Rewrite the equation as:

(x² - 4)(dy)/(dx) - 2xy = -2x(x² - 4)²

Dividing by $(x² - 4)²$ $(x^2 - 4)^2$ to create an exact differential:

((x² - 4)y' - 2xy)/((x² - 4)²) = -2x

Observe that the LHS is the exact derivative of $(y)/(x² - 4)$ $\frac{y}{x^2 - 4}$:

(d)/(dx)((y)/(x² - 4)) = -2x

Integrating both sides:

(y)/(x² - 4) = -x² + C

y = (-x² + C)(x² - 4)

Step 1: Finding C

The curve passes through $(3, 15)$ $(3, 15)$. Substitute $x=3, y=15$ $x=3, y=15$:

15 = (-(3)² + C)(3² - 4)

15 = (-9 + C)(5) 3 = -9 + C C = 12

The function is $y = (-x² + 12)(x² - 4)$ $y = (-x^2 + 12)(x^2 - 4)$.

Step 2: Maxima Calculation

To find the local maximum, differentiate and set $y' = 0$ $y' = 0$:

y' = -2x(x² - 4) + 2x(-x² + 12) = -2x(x² - 4 + x² - 12) = -2x(2x² - 16) = -4x(x² - 8)

Critical points at $x = 0$ $x = 0$ and $x = \pm 2\sqrt(2)$ $x = \pm 2\sqrt{2}$. Given domain $x > 2$ $x > 2$, the valid critical point is $x = 2\sqrt(2)$ $x = 2\sqrt{2}$. Check double derivative or nature: It produces a maximum at $x = 2\sqrt(2)$ $x = 2\sqrt{2}$.

Substitute $x² = 8$ $x^2 = 8$ into the function for the maximum value:

ymax = (-8 + 12)(8 - 4) = (4)(4) = 16

Pattern Recognition

Any differential equation framed like $v u' - u v' = K$ $v u' - u v' = K$ hints directly at quotient rule exact differentials. Identifying this skips the entire linear integrating factor calculation.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Application of Derivatives

Q21 jee_main_2026_24_january_morning Integral Equations

Let a differentiable function

f

satisfy the equation

\int₀³⁶f((tx)/(36))dt = 4α f(x)

. If

y = f(x)

is a standard parabola passing through the points

(2, 1)

and

(-4, β)

, Then

βα

is equal to

Numerical Answer. Answer: 64 to 64

Solution

Related Formula

Newton-Leibniz Rule for differentiating under the integral sign:

(d)/(dx) ( \intu(x)v(x) f(t) dt ) = f(v(x))\cdot v'(x) - f(u(x))\cdot u'(x)

Core Logic

Put (tx)/(36) = y \Rightarrow (x)/(36) dt = dy \Rightarrow dt = (36)/(x) dy

Limits: when $t = 0, y = 0$ $t = 0, y = 0$; when $t = 36, y = x$ $t = 36, y = x$.

\int₀^x f(y) (36)/(x) dy = 4α f(x)

\int₀^x f(y) dy = (α x)/(9) f(x)

Step 1: Differentiating with respect to x

f(x) = (α)/(9) ( f(x) + x f'(x) )

f(x) (1 - (α)/(9)) = (α x)/(9) f'(x)

(f'(x))/(f(x)) = ( (9)/(α) - 1 ) (1)/(x)

Step 2: Integration for Function

ln f(x) = ( (9)/(α) - 1 ) ln x + ln c

f(x) = c x((9)/(α) - 1)

Given $y = f(x)$ $y = f(x)$ is a standard parabola (typically $y² = 4ax$ $y^2 = 4ax$ or $x² = 4ay$ $x^2 = 4ay$). From the polynomial form, it implies degree 2.

(9)/(α) - 1 = 2 \Rightarrow (9)/(α) = 3 \Rightarrow α = 3

So, $f(x) = c x²$ $f(x) = c x^2$.

Step 3: Point Substitution

Passes through $(2, 1)$ $(2, 1)$:

1 = c(2)² \Rightarrow 4c = 1 \Rightarrow c = (1)/(4)

Thus, $y = (x²)/(4)$ $y = \frac{x^2}{4}$. Passes through $(-4, β)$ $(-4, \beta)$:

β = ((-4)²)/(4) = (16)/(4) = 4

β^α = 4³ = 64

Pattern Recognition

Integral equations of the form $\int f(tx)dt$ $\int f(tx)dt$ inherently resolve by standardizing the boundary variable ( $tx \to y$ $tx \to y$) and differentiating, revealing an underlying homogeneous differential equation.

Chapter Mix

Class 12 Maths: Differential Equations Class 11 Maths: Conic Sections

Solution & Explanation

Related Formula

Core Logic

Step 1: Execute Integration

Step 2: Apply Boundary Conditions

Step 3: Evaluate at x = e

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Determine Constant C

Step 2: Calculate for x = 49

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integrate both sides

Step 2: Find Integration Constant

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve the Differential Equation

Step 2: Calculate Sequence Values

Step 3: Calculate the Mean

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding C

Step 2: Maxima Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Differentiating with respect to x

Step 2: Integration for Function

Step 3: Point Substitution

Pattern Recognition

Chapter Mix

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