Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = 2(y + 2 \sin x - 5)x - 2 \cos x$ such that $y(0) = 7$. Then $y(\pi)$ is equal to : Options: (A) $2e^{\pi^2} + 5$ (B) $e^{\pi^2} + 5$ (C) $3e^{\pi^2} + 5$ (D) $7e^{\pi^2} + 5$

### Analyzing the Setup We are given the differential equation: $$\frac{dy}{dx} = 2(y + 2 \sin x - 5)x - 2 \cos x$$ with the initial condition $y(0) = 7$. Our mission is to determine the value of $y(\pi)$. ### Unmasking the Linear Form First, we expand the right-hand side of the equation: $$\frac{dy}{dx} = 2xy + 4x \sin x - 10x - 2 \cos x$$ Next, we rearrange the terms to isolate the linear structure by bringing the $y$ term to the left side: $$\frac{dy}{dx} - 2xy = 2x(2 \sin x - 5) - 2 \cos x$$ This is a classic **Linear Differential Equation** of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -2x$ and $Q(x) = 2x(2 \sin x - 5) - 2 \cos x$. ### The Magic of the Integrating Factor To solve this, we calculate the Integrating Factor ($IF$): $$IF = e^{\int P(x) dx} = e^{\int -2x dx} = e^{-x^2}$$ Multiplying the entire differential equation by $e^{-x^2}$ allows us to express the left side as a single derivative: $$\frac{d}{dx} \left( y \cdot e^{-x^2} \right) = e^{-x^2} \left[ 2x(2 \sin x - 5) - 2 \cos x \right]$$ ### The 'Aha!' Moment Integrating both sides with respect to $x$, we obtain: $$y \cdot e^{-x^2} = \int e^{-x^2} \left[ 2x(2 \sin x - 5) - 2 \cos x \right] dx + C$$ By observing the structure of the integrand, we recognize it as the result of the product rule applied to $e^{-x^2}(5 - 2 \sin x)$. Specifically: $$\frac{d}{dx} \left[ e^{-x^2}(5 - 2 \sin x) \right] = e^{-x^2}(-2 \cos x) + (5 - 2 \sin x)e^{-x^2}(-2x)$$ Factoring out $e^{-x^2}$ yields the original integrand. Thus, the integral simplifies to: $$y \cdot e^{-x^2} = e^{-x^2}(5 - 2 \sin x) + C$$ ### The Final Stretch Multiplying through by $e^{x^2}$ isolates $y$: $$y(x) = 5 - 2 \sin x + C e^{x^2}$$ Using the initial condition $y(0) = 7$: $$7 = 5 - 2 \sin(0) + C e^0 \Rightarrow 7 = 5 + C \Rightarrow C = 2$$ The particular solution is $y(x) = 5 - 2 \sin x + 2 e^{x^2}$. Finally, substituting $x = \pi$: $$y(\pi) = 5 - 2 \sin(\pi) + 2 e^{\pi^2}$$ Since $\sin(\pi) = 0$, the final result is: **$y(\pi) = 2e^{\pi^2} + 5$**

JEE Main 2021 (27 August Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$ such that $y (0) = 7$ . Then $y (π)$ is equal to :

Select Answer:

Progressive Hints

Hint 1

The Spark: Do not be overwhelmed by the complexity of the expression. Look at the structure of the differential equation. It is a linear differential equation in the form

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

. Your first task is to rearrange the terms to isolate

\frac{d y}{d x}

and the

y

term on one side.

Hint 2

The Arsenal: Once you have the equation in the standard form

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

, you need to find the Integrating Factor (

I F

). Recall that

I F = e^{\int P (x) d x}

. This is the key to transforming the left side into the derivative of a product.

Hint 3

The Execution: After finding the

I F

, the general solution is

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

. Before you attempt to integrate the right side, look for a pattern. The integrand is likely the derivative of a product of

e^{- x^{2}}

and another function. Check the derivative of

e^{- x^{2}} (5 - 2 sin x)

Visualized Solution

The Given Equation & Identifying the Type

Given differential equation: $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$
Target: Find the value of $y (π)$ given $y (0) = 7$
Standard form of a Linear Differential Equation (LDE): $\frac{d y}{d x} + P (x) y = Q (x)$

Rearranging to Standard Form

Expanding RHS: $\frac{d y}{d x} = 2 x y + 4 x sin x - 10 x - 2 cos x$
Rearranging: $\frac{d y}{d x} - 2 x y = 2 x (2 sin x - 5) - 2 cos x$
Comparing with standard form: $P (x) = - 2 x$ and $Q (x) = 2 x (2 sin x - 5) - 2 cos x$

Calculating the Integrating Factor ( $I F$ )

Formula for Integrating Factor ( $I F$ ): $I F = e^{\int P (x) d x}$
Substitute $P (x) = - 2 x$ : $I F = e^{\int - 2 x d x}$
Evaluating the integral: $I F = e^{- x^{2}}$

The General Solution Setup

The general solution formula: $y \cdot (I F) = \int Q (x) \cdot (I F) d x + C$
Substituting $I F$ and $Q (x)$ : $y \cdot e^{- x^{2}} = \int e^{- x^{2}} [2 x (2 sin x - 5) - 2 cos x] d x + C$

Spotting the Exact Differential

Integrand: $e^{- x^{2}} [2 x (2 sin x - 5) - 2 cos x]$
This looks like the result of a product rule differentiation.
Let's check the derivative of $e^{- x^{2}} (5 - 2 sin x)$ using $\frac{d}{d x} [u \cdot v] = u^{'} v + u v^{'}$

Verifying the Derivative

$\frac{d}{d x} [e^{- x^{2}} (5 - 2 sin x)] = e^{- x^{2}} (- 2 cos x) + (5 - 2 sin x) e^{- x^{2}} (- 2 x)$
$= e^{- x^{2}} [- 2 cos x - 10 x + 4 x sin x]$
$= e^{- x^{2}} [2 x (2 sin x - 5) - 2 cos x]$
This perfectly matches our integrand!

Integrating and Solving for $y (x)$

Integrating both sides: $y \cdot e^{- x^{2}} = e^{- x^{2}} (5 - 2 sin x) + C$
Multiply by $e^{x^{2}}$ to isolate $y$ : $y (x) = 5 - 2 sin x + C e^{x^{2}}$

Applying Initial Condition to find $C$

Use the initial condition $y (0) = 7$ : $7 = 5 - 2 sin (0) + C e^{0^{2}}$
Since $sin (0) = 0$ and $e^{0} = 1$ : $7 = 5 - 0 + C (1)$
Solving for $C$ : $C = 2$
Particular solution: $y (x) = 5 - 2 sin x + 2 e^{x^{2}}$

Finding $y (π)$ (Final Result)

Substitute $x = π$ : $y (π) = 5 - 2 sin (π) + 2 e^{π^{2}}$
Since $sin (π) = 0$ : $y (π) = 5 - 0 + 2 e^{π^{2}}$
Final Answer: $y (π) = 2 e^{π^{2}} + 5$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

We are given the differential equation:

\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x

with the initial condition

y (0) = 7

. Our mission is to determine the value of

y (π)

Unmasking the Linear Form

First, we expand the right-hand side of the equation:

\frac{d y}{d x} = 2 x y + 4 x sin x - 10 x - 2 cos x

Next, we rearrange the terms to isolate the linear structure by bringing the

y

term to the left side:

\frac{d y}{d x} - 2 x y = 2 x (2 sin x - 5) - 2 cos x

This is a classic Linear Differential Equation of the form

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

, where

P (x) = - 2 x

and

Q (x) = 2 x (2 sin x - 5) - 2 cos x

The Magic of the Integrating Factor

To solve this, we calculate the Integrating Factor (

I F

I F = e^{\int P (x) d x} = e^{\int - 2 x d x} = e^{- x^{2}}

Multiplying the entire differential equation by

e^{- x^{2}}

allows us to express the left side as a single derivative:

\frac{d}{d x} (y \cdot e^{- x^{2}}) = e^{- x^{2}} [2 x (2 sin x - 5) - 2 cos x]

The 'Aha!' Moment

Integrating both sides with respect to

x

, we obtain:

y \cdot e^{- x^{2}} = \int e^{- x^{2}} [2 x (2 sin x - 5) - 2 cos x] d x + C

By observing the structure of the integrand, we recognize it as the result of the product rule applied to

e^{- x^{2}} (5 - 2 sin x)

. Specifically:

\frac{d}{d x} [e^{- x^{2}} (5 - 2 sin x)] = e^{- x^{2}} (- 2 cos x) + (5 - 2 sin x) e^{- x^{2}} (- 2 x)

Factoring out

e^{- x^{2}}

yields the original integrand. Thus, the integral simplifies to:

y \cdot e^{- x^{2}} = e^{- x^{2}} (5 - 2 sin x) + C

The Final Stretch

Multiplying through by

e^{x^{2}}

isolates

y

y (x) = 5 - 2 sin x + C e^{x^{2}}

Using the initial condition

y (0) = 7

7 = 5 - 2 sin (0) + C e^{0} \Rightarrow 7 = 5 + C \Rightarrow C = 2

The particular solution is

y (x) = 5 - 2 sin x + 2 e^{x^{2}}

. Finally, substituting

x = π

y (π) = 5 - 2 sin (π) + 2 e^{π^{2}}

Since

sin (π) = 0

, the final result is:

$y (π) = 2 e^{π^{2}} + 5$

Common Traps & FAQs

Why can't I just use separation of variables here?

Separation of variables only works if you can write the equation as

f (y) d y = g (x) d x

. Here, the

y

term is multiplied by

x

and added to other functions of

x

, making it impossible to isolate

y

and

x

completely. This is a classic Linear Differential Equation, and the Integrating Factor method is the standard, robust tool for this structure.

How do I know if the integral on the right side is solvable?

In JEE Advanced problems, if an integral looks impossible to solve using standard techniques like integration by parts or substitution, it is almost always an 'exact differential' in disguise. Always check if the integrand is the result of the product rule

\frac{d}{d x} (u \cdot v) = u^{'} v + u v^{'}

. In this case, the structure

e^{- x^{2}} \cdot (e x t so m e t hin g)

is a huge hint that it came from the product rule.

Why do we need the initial condition

y (0) = 7

The general solution of a first-order differential equation always contains an arbitrary constant

C

. This constant represents a family of curves. The initial condition

y (0) = 7

allows us to 'pin down' the specific curve from that family that passes through the point

(0, 7)

, giving us the unique solution.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$ such that $y (0) = 7$ . Then $y (π)$ is equal to :

Select Answer:

Visualized Solution

The Given Equation & Identifying the Type

Rearranging to Standard Form

Calculating the Integrating Factor ( $I F$ )

The General Solution Setup

Spotting the Exact Differential

Verifying the Derivative

Integrating and Solving for $y (x)$

Applying Initial Condition to find $C$

Finding $y (π)$ (Final Result)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Unmasking the Linear Form

The Magic of the Integrating Factor

The 'Aha!' Moment

The Final Stretch

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $sec x \frac{d y}{d x} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2}), y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ 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 $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Let the solution $y = y (x)$ of the differential equation $\frac{d y}{d x} - y = 1 + 4 sin x$ satisfy $y (π) = 1$ . Then $y (\frac{π}{2}) + 10$ 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 $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is

Select Answer:

Visualized Solution

The Given Equation & Identifying the Type

Rearranging to Standard Form

Calculating the Integrating Factor ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IF)

The General Solution Setup

Spotting the Exact Differential

Verifying the Derivative

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

Applying Initial Condition to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Unmasking the Linear Form

The Magic of the Integrating Factor

The 'Aha!' Moment

The Final Stretch

Similar Questions

.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 secxdxdy​−2y=2+3sinx,x∈(−2π​,2π​),y(0)=−47​. Then y(6π​) is equal to :

.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​+2y=sin(2x),y(0)=43​, then y(8π​) is equal to:

.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)sec2x,x∈(−π/2,π/2), such that y(0)=0, then y(−π/4) is equal to :

.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 dxdy​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=45​. Then 12(y(4π​)−e−2) is equal to _______.

.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 cosxdxdy​+2ysinx=sin2x, x∈(0,2π​). If y(π/3)=0, then y(π/4) is equal to :

.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 xdy=(y+x3cosx)dx with y(π)=0, then y(2π​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the solution y=y(x) of the differential equation dxdy​−y=1+4sinx satisfy y(π)=1. Then y(2π​)+10 is equal to ______

.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 :

.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 sinxdxdy​+ycosx=4x,x∈(0,π). If y(π/2)=0, then y(π/6) is equal to :

.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 (1+x2)dxdy​+y=etan−1x,y(1)=0. Then y(0) is

Calculating the Integrating Factor ( $I F$ )

Integrating and Solving for $y (x)$

Applying Initial Condition to find $C$

Finding $y (π)$ (Final Result)

Let $y = y (x)$ be the solution of the differential equation $sec x \frac{d y}{d x} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2}), y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ 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 $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Let the solution $y = y (x)$ of the differential equation $\frac{d y}{d x} - y = 1 + 4 sin x$ satisfy $y (π) = 1$ . Then $y (\frac{π}{2}) + 10$ 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 $y = y (x)$ be the solution of the differential equation $sin x \frac{d y}{d x} + y cos x = 4 x, x \in (0, π)$ . If $y (π /2) = 0$ , then $y (π /6)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} + y = e^{t a n^{- 1} x}, y (1) = 0$ . Then $y (0)$ is