Let y = y(x) be a differentiable function in the interval (0, infty) such that y(1) = 2 and x ( (t2 y(x)

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More Differential Equations Previous-Year Questions — Page 3

Q20 jee_main_2026_28_january_morning Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

x (dy)/(dx) - 2y = x³(2 - x³) ² y

x \neq 0

. If

y(2) = 0

, then

(y(1))

is equal to

A. $(3)/(4)$ $\frac{3}{4}$
B. $(7)/(4)$ $\frac{7}{4}$
C. $-(7)/(4)$ $-\frac{7}{4}$
D. $-(3)/(4)$ $-\frac{3}{4}$

Solution

Core Logic

Given differential equation:

x (dy)/(dx) - 2y = x³(2 - x³) ² y

Divide both sides by $² y$ $\cos^2 y$:

x ² y (dy)/(dx) - (2 y y)/( ² y) = x³(2 - x³)

x ² y (dy)/(dx) - 2 y = x³(2 - x³)

Divide by $x$ $x$:

² y (dy)/(dx) - (2)/(x) y = x²(2 - x³)

Step 1: Reduce to Linear Form

Let $y = t$ $\tan y = t$. Then $² y (dy)/(dx) = (dt)/(dx)$ . The equation becomes a standard Linear Differential Equation:

(dt)/(dx) - ((2)/(x)) t = x²(2 - x³)

Step 2: Integrating Factor

Integrating Factor (I.F.):

I.F. = e\int -(2)/(x) dx = e-2 ln x = (1)/(x²)

Solution of the LDE:

t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C

(t)/(x²) = \int (1)/(x²) \cdot x²(2 - x³) dx + C

( y)/(x²) = \int (2 - x³) dx + C

( y)/(x²) = 2x - (x⁴)/(4) + C

Step 3: Apply Boundary Conditions

We are given $y(2) = 0$ $y(2) = 0$ (so $(0) = 0$ $\tan(0) = 0$). Substitute $x = 2$ $x = 2$ and $y = 0$ $y = 0$:

0 = 2(2) - (2⁴)/(4) + C 0 = 4 - 4 + C C = 0

The specific solution is:

y = x² (2x - (x⁴)/(4)) = 2x³ - (x⁶)/(4)

Step 4: Evaluate tan(y(1))

Substitute $x = 1$ $x = 1$:

(y(1)) = 2(1)³ - (1⁶)/(4) = 2 - (1)/(4) = (7)/(4)

Chapter Mix

Class 12 Mathematics: Differential Equations

Q14 jee_main_2026_28_january_evening Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

x (dy)/(dx) - y = x² x, x in (0, π)

. If

y((π)/(2))=(π)/(2)

, then

6y((π)/(6))-8y((π)/(4))

is equal to :

A. $3π$ $3\pi$
B. $-3π$ $-3\pi$
C. $-π$ $-\pi$
D. $π$ $\pi$

Solution

Related Formula

d((y)/(x)) = (x dy - y dx)/(x²)

Core Logic

Rewrite the differential equation:

x dy - y dx = x² x dx

Divide by $x²$ $x^2$ to create an exact differential:

(x dy - y dx)/(x²) = x dx

d((y)/(x)) = x dx

Execution

Integrate both sides:

\int d((y)/(x)) = \int x dx

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

Given $y((π)/(2)) = (π)/(2)$ :

(π/2)/(π/2) = ln| (π)/(2)| + C \Rightarrow 1 = 0 + C \Rightarrow C = 1

Equation of curve: $y = x(ln| x| + 1)$ $y = x(\ln|\sin x| + 1)$

Evaluate at limits: $y((π)/(6)) = (π)/(6)(ln(1)/(2) + 1) = (π)/(6)(-ln 2 + 1)$ $y((π)/(4)) = (π)/(4)(ln 1\sqrt(2) + 1) = (π)/(4)(-(1)/(2)ln 2 + 1)$

Calculate $6y((π)/(6)) - 8y((π)/(4))$ :

= 6[(π)/(6)(-ln 2 + 1)] - 8[(π)/(4)(-(1)/(2)ln 2 + 1)]

= π(-ln 2 + 1) - 2π(-(1)/(2)ln 2 + 1)

= -πln 2 + π + πln 2 - 2π = -π

Pattern Recognition

Whenever you see $x dy - y dx$ $x \, dy - y \, dx$, immediately test division by $x²$ $x^2$, $y²$ $y^2$, or $xy$ $xy$ to convert it directly into exact differential forms like $d(y/x)$ $d(y/x)$ or $d(x/y)$ $d(x/y)$.

Chapter Mix

Class 12 Maths: Differential Equations

Q52 jee_main_2025_02_april_evening Linear Differential Equations

Let

f:[1,\infty) \to [2,\infty)

be a differentiable function. If

10\int₁xf(t)dt = 5xf(x) - x⁵ - 9

for all

x \geq 1

, then the value of

f(3)

is:

A. 18
B. 32
C. 22
D. 26

Solution

Related Formula

Leibniz Rule for Differentiation under Integral Sign: (d)/(dx) \intu(x)v(x) f(t) dt = f(v(x)) v'(x) - f(u(x)) u'(x)

Standard Linear Differential Equation: (dy)/(dx) + P(x) y = Q(x)

Core Logic

Differentiating both sides with respect to $x$ $x$ eliminates the definite integral and leads to a first-order linear differential equation.

Step 1: Differentiate both sides

Apply differentiation with respect to $x$ $x$ using the Leibniz Rule on the left side, and product rule on the right side:

10 f(x) = 5 f(x) + 5x f'(x) - 5x⁴

5 f(x) + 5x⁴ = 5x f'(x) f(x) + x⁴ = x f'(x)

Letting $y = f(x)$ $y = f(x)$, we rewrite it as:

(dy)/(dx) - (y)/(x) = x³

Step 2: Solve the Linear Differential Equation

The integrating factor (I.F.) is:

I.F. = e\int -(1)/(x) dx = e-ln x = (1)/(x)

Multiply by the I.F. and integrate:

y \cdot (1)/(x) = \int x³ \cdot (1)/(x) dx = \int x² dx = (x³)/(3) + C

Thus, the general solution is:

f(x) = (x⁴)/(3) + C x

Step 3: Apply Boundary Conditions

Substitute $x = 1$ $x = 1$ into the original integral equation:

10 \int₁¹ f(t) dt = 5(1) f(1) - 1⁵ - 9

0 = 5 f(1) - 10 f(1) = 2

Now, substitute $x = 1$ $x = 1$ and $f(1) = 2$ $f(1) = 2$ into our general solution to find $C$ $C$:

2 = (1)/(3) + C C = (5)/(3)

Therefore, the complete function is:

f(x) = (x⁴)/(3) + (5x)/(3)

Step 4: Compute f(3)

Evaluate the function at $x = 3$ $x = 3$:

f(3) = (3⁴)/(3) + (5(3))/(3) = 27 + 5 = 32

Pattern Recognition

Whenever a definite integral is defined from a constant to the variable $x$ $x$ within an equation, differentiating immediately reduces it to a differential equation. Finding the value of $f(a)$ $f(a)$ at the lower bound is a standard method to get the constant of integration.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q71 jee_main_2025_02_april_evening First Order Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

(dy)/(dx) + 2y ² x = 2 ² x + 3 x \cdot ² x

such that

y(0) = (5)/(4)

. Then

12(y((π)/(4)) - e⁻²)

is equal to ____________.

Numerical Answer. Answer: 21 to 21

Solution

Related Formula

Linear Differential equation form: (dy)/(dx) + P(x) y = Q(x)

Integrating Factor: I.F. = e\int P(x) dx

General solution: y \cdot I.F. = \int Q(x) \cdot I.F. dx + C

Core Logic

This is a first-order linear differential equation. We calculate the Integrating Factor first to write down the integral solution.

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

Here, $P(x) = 2 ² x$ $P(x) = 2\sec^2 x$ and $Q(x) = 2 ² x + 3 x \cdot ² x$ :

I.F. = e\int 2 ² x dx = e2 x

Step 2: Obtain the General Solution

Multiply both sides by the integrating factor:

y \cdot e2 x = \int e2 x ( 2 ² x + 3 x \cdot ² x ) dx

Let $t = x dt = ² x dx$ $t = \tan x \implies dt = \sec^2 x dx$. The integral becomes:

\int e2t (2 + 3t) dt = \int 2 e2t dt + 3 \int t e2t dt

Applying integration by parts for the second term:

3 \int t e2t dt = 3 [ t e2t2 - \int e2t2 dt ] = 3t e2t2 - 3e2t4

Summing all parts:

y \cdot e2 x = e2t + 3t e2t2 - 3e2t4 + C = e2 x [ 1 + (3 x)/(2) - (3)/(4) ] + C

y = (3 x)/(2) + (1)/(4) + C e-2 x

Step 3: Apply the boundary conditions

Using the initial boundary condition $y(0) = (5)/(4)$ $y(0) = \frac{5}{4}$:

(5)/(4) = (3(0))/(2) + (1)/(4) + C e⁰ C = 1

Thus, the complete function is:

y(x) = (3 x)/(2) + (1)/(4) + e-2 x

Step 4: Compute the final value

Evaluate the function at $x = (π)/(4)$ $x = \frac{\pi}{4}$:

y((π)/(4)) = (3(1))/(2) + (1)/(4) + e⁻² = (7)/(4) + e⁻²

Now, calculate the requested value:

12 ( y((π)/(4)) - e⁻² ) = 12 ( (7)/(4) ) = 21

Pattern Recognition

Integration by parts substitution: When integrating terms of the form $\int eat P(t) dt$ $\int e^{at} P(t) dt$ (where $P(t)$ $P(t)$ is a polynomial), substituting the polynomial variable directly simplifies the exponential integration factors cleanly.

Chapter Mix

Class 12 Mathematics: Differential Equations

Let y = y(x)$y = y(x)$ be a differentiable function in the interval (0, ∞)$(0, \infty)$ such that y(1) = 2$y(1) = 2$ and t → x ( (t² y(x) - x² y(t))/(x - t) ) = 3$\lim_{t \to x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3$ for each x > 0$x > 0$. Then 2y(2)$2y(2)$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Integrating Factor

Step 2: Solving the Differential Equation

Step 3: Using the Boundary Condition

Step 4: Evaluating the Target Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Differential Equations Previous-Year Questions — Page 3

Solution

Core Logic

Step 1: Reduce to Linear Form

Step 2: Integrating Factor

Step 3: Apply Boundary Conditions

Step 4: Evaluate tan(y(1))

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Differentiate both sides

Step 2: Solve the Linear Differential Equation

Step 3: Apply Boundary Conditions

Step 4: Compute f(3)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

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

Step 2: Obtain the General Solution

Step 3: Apply the boundary conditions

Step 4: Compute the final value

Pattern Recognition

Chapter Mix

More Differential Equations Questions — jee_main_2026_24_january_evening

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Let $y = y(x)$ $y = y(x)$ be a differentiable function in the interval $(0, \infty)$ $(0, \infty)$ such that $y(1) = 2$ $y(1) = 2$ and $t \to x ( (t² y(x) - x² y(t))/(x - t) ) = 3$ for each $x > 0$ $x > 0$. Then $2y(2)$ $2y(2)$ is equal to