JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: A function $y = f (x)$ satisfies $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$ with condition $f (0) = 0$ . Then $f (\frac{π}{2})$ is equal to

Q: Wait, why is the Integrating Factor $1 + \cos^2 x$ and not something else?

Remember the formula $IF = e^{\int P(x) dx}$. When you integrate $P(x) = -\frac{\sin 2x}{1 + \cos^2 x}$, you get $-\ln(1 + \cos^2 x)$. The negative sign is crucial! It makes the exponent $\ln((1 + \cos^2 x)^{-1})$, which simplifies to $\frac{1}{1 + \cos^2 x}$. Wait, let me recheck the math. Actually, the integral of $-\frac{\sin 2x}{1 + \cos^2 x} dx$ is $\ln(1 + \cos^2 x)$ because the derivative of $1 + \cos^2 x$ is $-2 \cos x \sin x = -\sin 2x$. So, the integral is $\int \frac{du}{u} = \ln u = \ln(1 + \cos^2 x)$. Therefore, $IF = e^{\ln(1 + \cos^2 x)} = 1 + \cos^2 x$. The math is correct!

Select Answer:

Visualized Solution

Identify the Differential Equation

Given equation: $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$
Boundary condition: $f (0) = 0$
Goal: Find the value of $f (\frac{π}{2})$

Rearrange to Standard Form

Rearranging the terms: $(1 + cos^{2} x) f^{'} (x) - (sin 2 x) f (x) = sin x$
Dividing by $(1 + cos^{2} x)$ : $f^{'} (x) - \frac{s i n 2 x}{1 + c o s ^{2} x} f (x) = \frac{s i n x}{1 + c o s ^{2} x}$

Identify $P (x)$ and $Q (x)$

Standard form: $\frac{d y}{d x} + P (x) y = Q (x)$
Comparing, we get:
$P (x) = - \frac{s i n 2 x}{1 + c o s ^{2} x}$
$Q (x) = \frac{s i n x}{1 + c o s ^{2} x}$

Calculate the Integrating Factor

Integrating Factor $I F = e^{\int P (x) d x}$
$I F = e^{\int - \frac{s i n 2 x}{1 + c o s ^{2} x} d x}$

Solve the Integral for IF

Let $u = 1 + cos^{2} x$
Differentiating: $d u = - 2 cos x sin x d x = - sin 2 x d x$
The integral becomes: $\int \frac{1}{u} d u = ln u$
Substituting back: $ln (1 + cos^{2} x)$

Simplify the Integrating Factor

$I F = e^{l n (1 + c o s^{2} x)}$
Using $e^{l n A} = A$ :
$I F = 1 + cos^{2} x$

Write the General Solution

General Solution: $f (x) \cdot I F = \int Q (x) \cdot I F d x + C$
$f (x) (1 + cos^{2} x) = \int \frac{s i n x}{1 + c o s ^{2} x} (1 + cos^{2} x) d x + C$
$f (x) (1 + cos^{2} x) = \int sin x d x + C$

Integrate and Add Constant

$f (x) (1 + cos^{2} x) = - cos x + C$

Apply Boundary Condition

At $x = 0, f (0) = 0$ :
$0 \cdot (1 + cos^{2} 0) = - cos 0 + C$
$0 = - 1 + C \Rightarrow C = 1$

Find the Final Value

The function is: $f (x) (1 + cos^{2} x) = 1 - cos x$
At $x = \frac{π}{2}$ :
$f (\frac{π}{2}) (1 + cos^{2} \frac{π}{2}) = 1 - cos \frac{π}{2}$
$f (\frac{π}{2}) (1 + 0) = 1 - 0$
$f (\frac{π}{2}) = 1$

Conclusion and Summary

Key Takeaway:
1. Convert to standard form $\frac{d y}{d x} + P y = Q$ .
2. Calculate $I F = e^{\int P d x}$ .
3. Use boundary conditions to find the constant $C$ .
Final Answer: $f (\frac{π}{2}) = 1$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

We are given the differential equation:

f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0

with the boundary condition

f (0) = 0

. Our goal is to determine the value of

f (\frac{π}{2})

To reveal the standard form of a first-order linear differential equation, we rearrange the terms:

(1 + cos^{2} x) f^{'} (x) - (sin 2 x) f (x) = sin x

Dividing the entire equation by

(1 + cos^{2} x)

to normalize the coefficient of

f^{'} (x)

, we obtain:

f^{'} (x) - \frac{sin 2 x}{1 + cos ^{2} x} f (x) = \frac{sin x}{1 + cos ^{2} x}

This matches the standard form

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

, where:

P (x) = - \frac{sin 2 x}{1 + cos ^{2} x}, Q (x) = \frac{sin x}{1 + cos ^{2} x}

The Hero of the Story

The Integrating Factor

To solve this, we calculate the Integrating Factor (

I F

) using the formula

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

First, we evaluate the integral of

P (x)

\int P (x) d x = \int - \frac{sin 2 x}{1 + cos ^{2} x} d x

Using the substitution

u = 1 + cos^{2} x

, we find

d u = - 2 cos x sin x d x = - sin 2 x d x

. The integral simplifies to:

\int \frac{1}{u} d u = ln u = ln (1 + cos^{2} x)

Thus, the Integrating Factor is:

I F = e^{l n (1 + c o s^{2} x)} = 1 + cos^{2} x

The Final Stretch

Integration and the Boundary Condition

Multiplying the normalized differential equation by the

I F

, the left side becomes the derivative of the product

f (x) \cdot I F

f (x) (1 + cos^{2} x) = \int Q (x) \cdot I F d x + C

Substituting the known values:

f (x) (1 + cos^{2} x) = \int \frac{sin x}{1 + cos ^{2} x} (1 + cos^{2} x) d x + C

The terms cancel out, leaving a simple integral:

f (x) (1 + cos^{2} x) = \int sin x d x + C = - cos x + C

Applying the boundary condition

f (0) = 0

f (0) (1 + cos^{2} 0) = - cos 0 + C \Rightarrow 0 (2) = - 1 + C \Rightarrow C = 1

The specific solution is:

f (x) (1 + cos^{2} x) = 1 - cos x

Finally, substituting

x = \frac{π}{2}

to find

f (\frac{π}{2})

f (\frac{π}{2}) (1 + cos^{2} \frac{π}{2}) = 1 - cos \frac{π}{2}

f (\frac{π}{2}) (1 + 0) = 1 - 0

The final result is: $f (\frac{π}{2}) = 1$

Common Traps & FAQs

Why did we divide by

(1 + cos^{2} x)

? Can't we just integrate directly?

That is a great question! If you try to integrate directly, you will find that

f (x)

and

f^{'} (x)

are mixed in a way that does not allow for simple separation of variables. By dividing, we force the equation into the standard linear form, which unlocks the powerful tool of the Integrating Factor, allowing us to solve it systematically.

Wait, why is the Integrating Factor

1 + cos^{2} x

and not something else?

Remember the formula

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

. When you integrate

P (x) = - \frac{s i n 2 x}{1 + c o s ^{2} x}

, you get

- ln (1 + cos^{2} x)

. The negative sign is crucial! It makes the exponent

ln ((1 + cos^{2} x)^{- 1})

, which simplifies to

\frac{1}{1 + c o s ^{2} x}

. Wait, let me recheck the math. Actually, the integral of

- \frac{s i n 2 x}{1 + c o s ^{2} x} d x

ln (1 + cos^{2} x)

because the derivative of

1 + cos^{2} x

- 2 cos x sin x = - sin 2 x

. So, the integral is

\int \frac{d u}{u} = ln u = ln (1 + cos^{2} x)

. Therefore,

I F = e^{l n (1 + c o s^{2} x)} = 1 + cos^{2} x

. The math is correct!

Why do we need the boundary condition

f (0) = 0

The general solution to a differential equation always includes an arbitrary constant

C

. Without the boundary condition, we would have a family of curves. The condition

f (0) = 0

acts like a pin, fixing our specific curve in the coordinate plane so we can find the exact value at

x = \frac{π}{2}

Animated Solution for Mathematics - Differential Equations: A function $y = f (x)$ satisfies $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$ with condition $f (0) = 0$ . Then $f (\frac{π}{2})$ is equal to

Select Answer:

Visualized Solution

Identify the Differential Equation

Rearrange to Standard Form

Identify $P (x)$ and $Q (x)$

Calculate the Integrating Factor

Solve the Integral for IF

Simplify the Integrating Factor

Write the General Solution

Integrate and Add Constant

Apply Boundary Condition

Find the Final Value

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Hero of the Story

The Final Stretch

Similar Questions

Let $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

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

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Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

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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 :

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If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A function y=f(x) satisfies f(x)sin2x+sinx−(1+cos2x)f′(x)=0 with condition f(0)=0. Then f(2π​) is equal to

Select Answer:

Visualized Solution

Identify the Differential Equation

Rearrange to Standard Form

Identify .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)

Calculate the Integrating Factor

Solve the Integral for IF

Simplify the Integrating Factor

Write the General Solution

Integrate and Add Constant

Apply Boundary Condition

Find the Final Value

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Hero of the Story

The Final Stretch

Similar Questions

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Animated Solution for Mathematics - Differential Equations: A function $y = f (x)$ satisfies $f (x) sin 2 x + sin x - (1 + cos^{2} x) f^{'} (x) = 0$ with condition $f (0) = 0$ . Then $f (\frac{π}{2})$ is equal to

Identify $P (x)$ and $Q (x)$

Let $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

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

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ 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 $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

If $x = f (y)$ is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{t a n^{- 1} y}) \frac{d y}{d x} = 0,$ $y \in (- \frac{π}{2}, \frac{π}{2})$ with $f (0) = 1$ , then $f (\frac{1}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :