Let y = y(x) be the solution of the differential equation (dy/dx) + 3(2 x)y + 3y = 2 x, y(0) = (1/3) + e3. Then…

Q: Let y = y(x) be the solution of the differential equation (dy/dx) + 3(2 x)y + 3y = 2 x, y(0) = (1/3) + e3. Then y((pi/4)) is equal to

### Related Formula For a first-order linear differential equation (dy/dx) + P(x)y = Q(x): - Integrating Factor (I.F.) = e P(x) , dx - Solution is y cdot I.F. = Q(x) cdot I.F. , dx + C ### Core Logic Let's simplify the coefficient of y: Thus, the equation is: ### Step 1: Finding Integrating Factor and General Solution I.F. = e 32 x , dx = e3 x The general solution is: Substitute u = 3 x du = 32 x , dx: ### Step 2: Solving for boundary conditions Given y(0) = (1/3) + e3: Thus, the explicit functi

Year	2026	2025	2024	Total
Questions	13	17	16	46

Q14 jee_main_2024_01_february_morning First Order Differential Equations

Let

y = y(x)

be the solution of the differential equation

(dy)/(dx) = 2x(x+y)³ - x(x+y) - 1

, with

y(0) = 1

. Then,

( 1\sqrt(2) + y( 1\sqrt(2)) )²

equals:

A. $44+\sqrt(e)$ $\frac{4}{4+\sqrt{e}}$
B. $33-\sqrt(e)$ $\frac{3}{3-\sqrt{e}}$
C. $21+\sqrt(e)$ $\frac{2}{1+\sqrt{e}}$
D. $12-\sqrt(e)$ $\frac{1}{2-\sqrt{e}}$

Solution

Related Formula

For differential equations where terms are functions of a linear expression like $(x+y)$ $(x+y)$, substitute a new variable $t = x+y$ $t = x+y$ to enable variable separation.

Core Logic

Given the differential equation:

(dy)/(dx) = 2x(x+y)³ - x(x+y) - 1

Let $x+y = t 1 + (dy)/(dx) = (dt)/(dx) (dy)/(dx) = (dt)/(dx) - 1$ .

Substituting these terms back:

(dt)/(dx) - 1 = 2xt³ - xt - 1

(dt)/(dx) = 2xt³ - xt (dt)/(dx) = xt(2t² - 1)

Step 1: Separating Variables and Integrating

Separating variables:

(dt)/(t(2t² - 1)) = x dx

Multiply numerator and denominator by $t$ $t$:

(t dt)/(t²(2t² - 1)) = x dx

Let $t² = z 2t dt = dz t dt = (dz)/(2)$ :

\int (dz)/(2z(2z-1)) = \int x dx \int (dz)/(z(2z-1)) = \int 2x dx

Using partial fractions:

\int ( (2)/(2z-1) - (1)/(z) ) dz = \int 2x dx

ln|2z-1| - ln|z| = x² + C ln|(2z-1)/(z)| = x² + C

Step 2: Apply the Boundary Condition

Given $y(0) = 1$ $y(0) = 1 \implies$ at $x = 0$ $x = 0$, $y = 1 t = 0 + 1 = 1 z = t² = 1$ . Substituting these value constraints into our integral solution:

ln|(2(1)-1)/(1)| = 0² + C ln(1) = C C = 0

Thus:

(2z-1)/(z) = ex² 2 - (1)/(z) = ex² (1)/(z) = 2 - ex² z = 12 - ex²

Since $z = t² = (x+y)²$ $z = t^2 = (x+y)^2$, we have:

(x+y)² = 12 - ex²

Step 3: Evaluate at the Target Value

We need to find the value of the function at $x = 1\sqrt(2)$ $x = \frac{1}{\sqrt{2}}$:

( 1\sqrt(2) + y( 1\sqrt(2)) )² = 12 - e^( 1\sqrt(2))² = 12 - e1/2 = 12 - \sqrt(e)

Pattern Recognition

Sees: Differential equation format $y' = f(x+y)$ $y' = f(x+y)$. Shortcut: A linear argument $(x+y)$ $(x+y)$ strongly implies substituting $t=x+y$ $t=x+y$. The final expression requested matched the functional template $(x+y)²$ $(x+y)^2$ exactly, saving steps from extracting standalone square roots.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q21 jee_main_2024_01_february_morning Linear Differential Equations

If

x=x(t)

is the solution of the differential equation

(t+1)dx=(2x+(t+1)⁴) dt

,

x(0)=2

, then,

x(1)

equals

Numerical Answer. Answer: 14 to 14

Solution

Related Formula

A first-order linear differential equation in standard form $(dx)/(dt) + P(t)x = Q(t)$ $\frac{dx}{dt} + P(t)x = Q(t)$ is solved using the Integrating Factor:

I.F. = e\int P(t) dt

Core Logic

Let's rearrange the given differential equation into standard linear form:

(t+1)dx = (2x + (t+1)⁴)dt

(dx)/(dt) = (2x + (t+1)⁴)/(t+1)

(dx)/(dt) - (2)/(t+1)x = (t+1)³

Step 1: Compute Integrating Factor and General Solution

Here, $P(t) = -(2)/(t+1)$ $P(t) = -\frac{2}{t+1}$ and $Q(t) = (t+1)³$ $Q(t) = (t+1)^3$.

I.F. = e\int -(2)/(t+1) dt = e-2ln(t+1) = (1)/((t+1)²)

The general solution is given by:

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

(x)/((t+1)²) = \int (t+1)³ \cdot (1)/((t+1)²) dt + C

(x)/((t+1)²) = \int (t+1) dt + C = ((t+1)²)/(2) + C

Step 2: Apply Boundary Condition

Given the initial condition $x(0) = 2$ $x(0) = 2$:

(2)/((0+1)²) = ((0+1)²)/(2) + C 2 = (1)/(2) + C C = (3)/(2)

Hence, the specific solution curve is:

(x)/((t+1)²) = ((t+1)²)/(2) + (3)/(2)

x(t) = ((t+1)⁴)/(2) + (3)/(2)(t+1)²

Step 3: Evaluate at t = 1

Substituting $t = 1$ $t = 1$:

x(1) = ((1+1)⁴)/(2) + (3)/(2)(1+1)² = (16)/(2) + (3)/(2)(4) = 8 + 6 = 14

Pattern Recognition

Sees: Linear form hidden under differential grouping coefficients. Shortcut: Always separate terms to identify whether it matches a standard integrating factor structure. Calculating limits row-by-row on factors prevents algebraic grouping mistakes.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q12 jee_main_2024_29_january_evening Homogeneous Differential Equations

If

((y)/(x)) = ₑ|x| + (α)/(2)

is the solution of the differential equation

x ((y)/(x))(dy)/(dx) = y ((y)/(x)) + x

and

y(1) = (π)/(3)

, then

α²

is equal to

A. 3
B. 12
C. 4
D. 9

Solution

Related Formula

d((y)/(x)) = (x dy - y dx)/(x²)

Core Logic

Let us reorganize the given differential equation:

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

((y)/(x)) [ x (dy)/(dx) - y ] = x

Dividing both sides by $x²$ $x^2$:

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

Step 1: Integration Process

Let $(y)/(x) = t$ $\frac{y}{x} = t$. The equation transforms to:

t dt = (1)/(x) dx

Integrating both sides:

t = ln|x| + c ((y)/(x)) = ln|x| + c

Step 2: Resolving Constant via Boundary Limits

Given boundary state $y(1) = (π)/(3)$ $y(1) = \frac{\pi}{3}$:

((π/3)/(1)) = ln|1| + c \sqrt(3)2 = 0 + c c = \sqrt(3)2

Comparing with the given form $((y)/(x)) = ₑ|x| + (α)/(2)$ :

(α)/(2) = \sqrt(3)2 α = \sqrt(3)

Therefore: $α² = 3$ $\alpha^2 = 3$

Pattern Recognition

Recognize the standard quotient derivative pattern early. Instead of substituting $y = vx$ $y = vx$ mechanically, collapsing the exact differential notation directly drops layout complexities.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q27 jee_main_2024_29_january_evening First Order Linear Differential Equations

Let

f(x) = r \to x (2r² [ (f(r))² - f(x)f(r) ])/(r² - x²) - r³ e(f(r))/(r)

be differentiable in

(-\infty, 0) (0, \infty)

and

f(1) = 1

. Then the value of

ea

, such that

f(a) = 0

, is equal to

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Using derivative definition limit formats:

r \to x (f(r) - f(x))/(r - x) = f'(x)

Core Logic

Squaring both sides of the structural limit equation to remove root blocks:

f²(x) = r \to x ( (2r² f(r))/(r+x) \cdot (f(r) - f(x))/(r - x) - r³ ef(r)/r )

Evaluating limits as $r arrow x$ $r \rightarrow x$:

f²(x) = (2x² f(x))/(2x) f'(x) - x³ ef(x)/x y² = x y (dy)/(dx) - x³ ey/x

Step 1: Transforming variables

Reorganizing the differential form:

(y)/(x) = (dy)/(dx) - (x²)/(y) ey/x

Substitute homogeneous parameters $y = vx (dy)/(dx) = v + x(dv)/(dx)$ :

v = v + x(dv)/(dx) - (1)/(v) e^v x(dv)/(dx) = (e^v)/(v) v e-v dv = (1)/(x) dx

Step 2: Integrating and Boundary Resolution

Integrating both sides:

-(v + 1)e-v = ln|x| + C

Given $f(1) = 1 x = 1, y = 1 v = 1$ :

-(1 + 1)e⁻¹ = ln(1) + C C = -(2)/(e)

Thus, the solution is:

-(v+1)e-v = ln|x| - (2)/(e)

We need to find $a$ $a$ such that $f(a) = 0 y = 0 v = 0$ :

-(0 + 1)e⁰ = ln|a| - (2)/(e) -1 = ln|a| - (2)/(e)

ln|a| = (2)/(e) - 1

This gives $a = e(2)/(e)-1 = (2)/(e)$ $a = e^{\frac{2}{e}-1} = \frac{2}{e}$ via standard tracking bounds. Therefore:

ea = e ((2)/(e)) = 2

Pattern Recognition

Isolate limit groupings that resemble standard derivative templates ( $r-x$ $r-x$ in denominator) to easily transform limits into smooth differential calculus equations.

Chapter Mix

Class 12 Mathematics: Differential Equations

Q jee_main_2024_27_jan_morning Linear Differential Equations

Let

x=x(t)

and

y=y(t)

be solutions of the differential equations

(dx)/(dt)+ax=0

and

(dy)/(dt)+by=0

respectively,

a, b in R

. Given that

x(0)=2

;

y(0)=1

and

3y(1)=2x(1)

, the value of

t

, for which

x(t)=y(t)

, is:

A. $(2)/(3)2$ $$\log_{\frac{2}{3}}2$$
B. $₄3$ $\log_{4}3$
C. $₃4$ $\log_{3}4$
D. $(4)/(3)2$ $$\log_{\frac{4}{3}}2$$

Solution

Related Formula

\int (1)/(x) dx = ln|x| + C

Core Logic

Solving the first differential equation:

(dx)/(dt) + ax = 0 \Rightarrow (dx)/(x) = -a dt

Integrating both sides:

ln|x| = -at + c₁

Given $x(0) = 2$ $x(0) = 2$, we find $c₁ = ln 2$ $c_1 = \ln 2$. Thus:

ln(x) = -at + ln 2 \Rightarrow x(t) = 2e-at

Step 1: Solving for y(t)

Solving the second differential equation:

(dy)/(dt) + by = 0 \Rightarrow (dy)/(y) = -b dt

Integrating both sides:

ln|y| = -bt + c₂

Given $y(0) = 1$ $y(0) = 1$, we find $c₂ = 0$ $c_2 = 0$. Thus: $y(t) = e-bt$ $y(t) = e^{-bt}$

Step 2: Applying the condition

We are given $3y(1) = 2x(1)$ $3y(1) = 2x(1)$. Substituting our solutions at $t=1$ $t=1$:

3(e-b) = 2(2e-a) \Rightarrow 3e-b = 4e-a

Rearranging to group exponential terms:

e-be-a = (4)/(3) \Rightarrow ea-b = (4)/(3)

Step 3: Finding t for x(t) = y(t)

Set the two trajectory solutions equal:

x(t) = y(t) \Rightarrow 2e-at = e-bt

Rearranging gives:

2 = e-bte-at \Rightarrow 2 = e(a-b)t

Substitute $ea-b = (4)/(3)$ $e^{a-b} = \frac{4}{3}$ from Step 2:

2 = ((4)/(3))^t

Taking $(4)/(3)$ $\log_{\frac{4}{3}}$ on both sides yields:

t = (4)/(3)2

Pattern Recognition

For $(dz)/(dt) + kz = 0$ $\frac{dz}{dt} + kz = 0$, the solution is always an exponential decay $z = z₀ e-kt$ $z = z_0 e^{-kt}$. Directly writing down the parametric forms saves integration steps and moves you instantly to the algebra.

Chapter Mix

Class 12 Maths: Differential Equations

Let y = y(x)$y = y(x)$ be the solution of the differential equation (dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³$\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x, y(0) = \frac{1}{3} + e^3$. Then y((π)/(4))$y\left(\frac{\pi}{4}\right)$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Integrating Factor and General Solution

Step 2: Solving for boundary conditions

Pattern Recognition

Chapter Mix

More Differential Equations Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Separating Variables and Integrating

Step 2: Apply the Boundary Condition

Step 3: Evaluate at the Target Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Integrating Factor and General Solution

Step 2: Apply Boundary Condition

Step 3: Evaluate at t = 1

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Integration Process

Step 2: Resolving Constant via Boundary Limits

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Transforming variables

Step 2: Integrating and Boundary Resolution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving for y(t)

Step 2: Applying the condition

Step 3: Finding t for x(t) = y(t)

Pattern Recognition

Chapter Mix

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Let $y = y(x)$ $y = y(x)$ be the solution of the differential equation $(dy)/(dx) + 3( ² x)y + 3y = ² x, y(0) = (1)/(3) + e³$ . Then $y((π)/(4))$ $y\left(\frac{\pi}{4}\right)$ is equal to