JEE Main 2025 April

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Q: Why did we choose $u = \tan x$ for substitution?

Look at the integral: $\int (2 \sec^2 x + 3 \tan x \sec^2 x) e^{2 \tan x} dx$. We see $\tan x$ and its derivative $\sec^2 x$ appearing together. Whenever you see a function and its derivative in an integral, substitution is your best friend. It transforms a complex trigonometric integral into a simple algebraic one.

Q: What happens to the constant $C$?

The constant $C$ represents the family of all possible solutions to the differential equation. By using the initial condition $y(0) = 5/4$, we are essentially 'pinning' the curve to a specific point. This allows us to select the one unique member of that family that satisfies our specific problem.

Enter Numerical Value:

Visualized Solution

Identifying the Linear Form

The given equation is: $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x sec^{2} x$
This matches the standard form of a Linear Differential Equation (LDE):
$\frac{d y}{d x} + P (x) y = Q (x)$

Extracting $P (x)$ and $Q (x)$

By comparing with the standard form, we identify:
$P (x) = 2 sec^{2} x$
$Q (x) = 2 sec^{2} x + 3 tan x sec^{2} x$

Calculating the Integrating Factor

The Integrating Factor (IF) is given by: $I F = e^{\int P (x) d x}$
Substituting $P (x)$ : $I F = e^{\int 2 s e c^{2} x d x}$
Since $\int sec^{2} x d x = tan x$ :
$I F = e^{2 t a n x}$

Setting up the General Solution

The general solution is: $y \cdot I F = \int Q (x) \cdot I F d x$
Substituting the values:
$y \cdot e^{2 t a n x} = \int e^{2 t a n x} (2 sec^{2} x + 3 tan x sec^{2} x) d x$

The Power of Substitution

Let $u = tan x$
Differentiating both sides: $d u = sec^{2} x d x$
The integral transforms to:
$\int e^{2 u} (2 + 3 u) d u$

Splitting the Integral

Split the integral into two parts:
$I = \int 2 e^{2 u} d u + \int 3 u e^{2 u} d u$
Evaluating the first part:
$\int 2 e^{2 u} d u = 2 \cdot \frac{e ^{2 u}}{2} = e^{2 u}$

Integration by Parts

Using Integration by Parts for $\int 3 u e^{2 u} d u$ :
Let $f = 3 u$ , $g^{'} = e^{2 u}$
$\int 3 u e^{2 u} d u = 3 u (\frac{e ^{2 u}}{2}) - \int 3 (\frac{e ^{2 u}}{2}) d u$
$= \frac{3}{2} u e^{2 u} - \frac{3}{4} e^{2 u}$

Combining the Results

Total Integral $I = e^{2 u} + \frac{3}{2} u e^{2 u} - \frac{3}{4} e^{2 u} + C$
$I = \frac{3}{2} u e^{2 u} + \frac{1}{4} e^{2 u} + C$
Substitute back $u = tan x$ :
$y \cdot e^{2 t a n x} = \frac{3}{2} tan x e^{2 t a n x} + \frac{1}{4} e^{2 t a n x} + C$

Isolating $y (x)$

Divide by $e^{2 t a n x}$ to isolate $y$ :
$y = \frac{3}{2} tan x + \frac{1}{4} + C e^{- 2 t a n x}$

Using the Initial Condition

Given: $y (0) = \frac{5}{4}$
Substitute $x = 0$ and $y = \frac{5}{4}$ into the general solution:
$\frac{5}{4} = \frac{3}{2} tan (0) + \frac{1}{4} + C e^{- 2 t a n (0)}$
$\frac{5}{4} = 0 + \frac{1}{4} + C (1)$
$C = 1$

The Particular Solution

The particular solution is:
$y (x) = \frac{3}{2} tan x + \frac{1}{4} + e^{- 2 t a n x}$

Evaluating $y (\frac{π}{4})$

To find $y (\frac{π}{4})$ , substitute $x = \frac{π}{4}$ :
$y (\frac{π}{4}) = \frac{3}{2} tan (\frac{π}{4}) + \frac{1}{4} + e^{- 2 t a n (\frac{π}{4})}$
$y (\frac{π}{4}) = \frac{3}{2} (1) + \frac{1}{4} + e^{- 2}$
$y (\frac{π}{4}) = \frac{7}{4} + e^{- 2}$

Final Calculation

We need to find: $12 (y (\frac{π}{4}) - e^{- 2})$
Substitute $y (\frac{π}{4}) = \frac{7}{4} + e^{- 2}$ :
$12 (\frac{7}{4} + e^{- 2} - e^{- 2})$
$= 12 \cdot \frac{7}{4}$
$= 3 \cdot 7 = 21$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x sec^{2} x

This is a Linear Differential Equation of the form:

\frac{d y}{d x} + P (x) y = Q (x)

By comparing the terms, we identify:

P (x) = 2 sec^{2} x

Q (x) = 2 sec^{2} x + 3 tan x sec^{2} x

The Magic Multiplier

To solve this, we calculate the Integrating Factor (

I F

I F = e^{\int P (x) d x} = e^{\int 2 s e c^{2} x d x} = e^{2 t a n x}

This

I F

acts as the engine for our solution, allowing us to condense the left-hand side of the equation.

The Transformation

The general solution is given by the formula

y \cdot I F = \int Q (x) \cdot I F d x

. Substituting our known values:

y \cdot e^{2 t a n x} = \int (2 sec^{2} x + 3 tan x sec^{2} x) e^{2 t a n x} d x

To simplify, we use the substitution

u = tan x

, which implies

d u = sec^{2} x d x

. The integral becomes:

\int (2 + 3 u) e^{2 u} d u

Integration by Parts

We split the integral into two parts:

\int 2 e^{2 u} d u + \int 3 u e^{2 u} d u = e^{2 u} + (\frac{3}{2} u e^{2 u} - \frac{3}{4} e^{2 u}) + C

Simplifying the expression, we obtain:

y \cdot e^{2 t a n x} = \frac{3}{2} tan x e^{2 t a n x} + \frac{1}{4} e^{2 t a n x} + C

Dividing by

e^{2 t a n x}

, the general solution is:

y = \frac{3}{2} tan x + \frac{1}{4} + C e^{- 2 t a n x}

Final Calculation

We apply the initial condition

y (0) = \frac{5}{4}

. Since

tan (0) = 0

\frac{5}{4} = 0 + \frac{1}{4} + C (1) \Rightarrow C = 1

The particular solution is:

y (x) = \frac{3}{2} tan x + \frac{1}{4} + e^{- 2 t a n x}

Evaluating at

x = \frac{π}{4}

y (\frac{π}{4}) = \frac{3}{2} (1) + \frac{1}{4} + e^{- 2} = \frac{7}{4} + e^{- 2}

The final required value is:

12 (y (\frac{π}{4}) - e^{- 2}) = 12 (\frac{7}{4} + e^{- 2} - e^{- 2}) = 12 \cdot \frac{7}{4} = 21

Common Traps & FAQs

Why can't we just separate the variables directly?

Great question! Separation of variables only works if you can group all

y

terms with

d y

and all

x

terms with

d x

. Here, the term

2 y sec^{2} x

links

y

and

x

in a way that makes separation impossible. That is why we use the Linear Differential Equation method—it is designed specifically for this kind of coupling.

Why did we choose

u = tan x

for substitution?

Look at the integral:

\int (2 sec^{2} x + 3 tan x sec^{2} x) e^{2 t a n x} d x

. We see

tan x

and its derivative

sec^{2} x

appearing together. Whenever you see a function and its derivative in an integral, substitution is your best friend. It transforms a complex trigonometric integral into a simple algebraic one.

What happens to the constant

C

The constant

C

represents the family of all possible solutions to the differential equation. By using the initial condition

y (0) = 5/4

, we are essentially 'pinning' the curve to a specific point. This allows us to select the one unique member of that family that satisfies our specific problem.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Identifying the Linear Form

Extracting $P (x)$ and $Q (x)$

Calculating the Integrating Factor

Setting up the General Solution

The Power of Substitution

Splitting the Integral

Integration by Parts

Combining the Results

Isolating $y (x)$

Using the Initial Condition

The Particular Solution

Evaluating $y (\frac{π}{4})$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic Multiplier

The Transformation

Integration by Parts

Final Calculation

Similar Questions

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

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to:

Enter Numerical Value:

Visualized Solution

Identifying the Linear Form

Extracting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Calculating the Integrating Factor

Setting up the General Solution

The Power of Substitution

Splitting the Integral

Integration by Parts

Combining the Results

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

Using the Initial Condition

The Particular Solution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(4π​)

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic Multiplier

The Transformation

Integration by Parts

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation dxdy​+(tanx)y=sinx,0≤x≤3π​, with y(0)=0, then y(4π​) equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation cosec2xdy+2dx=(1+ycos2x)cosec2xdx, with y(4π​)=0. Then, the value of (y(0)+1)2 is equal to:

Extracting $P (x)$ and $Q (x)$

Isolating $y (x)$

Evaluating $y (\frac{π}{4})$

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

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to: