JEE Main 2021 (17 March Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Select Answer:

Progressive Hints

Hint 1

Look at the structure. It is a classic Linear Differential Equation in disguise. Your first mission is to isolate

\frac{d y}{d x}

by dividing both sides by the term

cos x (3 sin x + cos x + 3)

. Don't be scared by the complexity; once you separate the terms, the equation will reveal its true, simpler form.

Hint 2

Once you have the equation in the 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}

. In this case,

P (x) = - tan x

. Calculate this integral carefully—it simplifies beautifully!

Hint 3

After multiplying by the I.F., you will be left with the integral

\int \frac{1}{3 s i n x + c o s x + 3} d x

. This is a classic integral that screams for the Weierstrass substitution: let

t = tan (\frac{x}{2})

. Remember to replace

d x

with

\frac{2 d t}{1 + t ^{2}}

and express

sin x

and

cos x

in terms of

t

Visualized Solution

Analyze the Given Differential Equation

Given Equation: $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$
Initial Condition: $y (0) = 0$
Goal: Find the value of $y (\frac{π}{3})$

Rearranging for $\frac{d y}{d x}$

Dividing by $d x$ and $cos x (3 sin x + cos x + 3)$ :
$\frac{d y}{d x} = \frac{1 + y s i n x ( 3 s i n x + c o s x + 3 )}{c o s x ( 3 s i n x + c o s x + 3 )}$
Separating the terms:
$\frac{d y}{d x} = \frac{1}{c o s x ( 3 s i n x + c o s x + 3 )} + \frac{y s i n x ( 3 s i n x + c o s x + 3 )}{c o s x ( 3 s i n x + c o s x + 3 )}$

Identifying the Linear Form

Simplifying the second term:
$\frac{d y}{d x} = \frac{1}{c o s x ( 3 s i n x + c o s x + 3 )} + y tan x$
Rearranging to standard Linear Differential Equation (LDE) form:
$\frac{d y}{d x} - (tan x) y = \frac{1}{c o s x ( 3 s i n x + c o s x + 3 )}$
Comparing with $\frac{d y}{d x} + P (x) y = Q (x)$ :
$P (x) = - tan x$
$Q (x) = \frac{1}{c o s x ( 3 s i n x + c o s x + 3 )}$

Calculating the Integrating Factor (I.F.)

Integrating Factor (I.F.) formula: $I . F . = e^{\int P (x) d x}$
Substituting $P (x) = - tan x$ :
$I . F . = e^{\int - t a n x d x}$
Since $\int - tan x d x = ln ∣ cos x ∣$ :
$I . F . = e^{l n ∣ c o s x ∣} = cos x$

Setting Up the General Solution

General Solution: $y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x + C$
Substituting $I . F . = cos x$ and $Q (x)$ :
$y cos x = \int \frac{c o s x}{c o s x ( 3 s i n x + c o s x + 3 )} d x + C$
Cancelling $cos x$ :
$y cos x = \int \frac{1}{3 s i n x + c o s x + 3} d x + C$

Half-Angle Substitution Method

To solve $\int \frac{d x}{3 s i n x + c o s x + 3}$ , use substitution $t = tan (\frac{x}{2})$ :
$sin x = \frac{2 t}{1 + t ^{2}}$
$cos x = \frac{1 - t ^{2}}{1 + t ^{2}}$
$d x = \frac{2 d t}{1 + t ^{2}}$

Transforming the Integral to $t$

Substituting into the integral:
$\int \frac{\frac{2 d t}{1 + t ^{2}}}{3 ( \frac{2 t}{1 + t ^{2}} ) + ( \frac{1 - t ^{2}}{1 + t ^{2}} ) + 3}$
Multiplying numerator and denominator by $(1 + t^{2})$ :
$\int \frac{2 d t}{6 t + ( 1 - t ^{2} ) + 3 ( 1 + t ^{2} )}$

Simplifying the Quadratic Expression

Simplifying the denominator:
$\int \frac{2 d t}{6 t + 1 - t ^{2} + 3 + 3 t ^{2}} = \int \frac{2 d t}{2 t ^{2} + 6 t + 4}$
Dividing by $2$ :
$\int \frac{d t}{t ^{2} + 3 t + 2}$

Integrating via Partial Fractions

Factoring the denominator: $t^{2} + 3 t + 2 = (t + 1) (t + 2)$
Using Partial Fractions:
$\frac{1}{( t + 1 ) ( t + 2 )} = \frac{1}{t + 1} - \frac{1}{t + 2}$
Integrating:
$\int (\frac{1}{t + 1} - \frac{1}{t + 2}) d t = ln ∣ t + 1∣ - ln ∣ t + 2∣ = ln ∣ \frac{t + 1}{t + 2} ∣$

Applying the Initial Condition

General Solution in terms of $x$ :
$y cos x = ln ∣ \frac{t a n ( \frac{x}{2} ) + 1}{t a n ( \frac{x}{2} ) + 2} ∣ + C$
Using $y (0) = 0$ :
$0 \cdot cos (0) = ln ∣ \frac{t a n ( 0 ) + 1}{t a n ( 0 ) + 2} ∣ + C$
$0 = ln (\frac{1}{2}) + C \Rightarrow C = - ln (\frac{1}{2}) = ln 2$

The Particular Solution

Substituting $C = ln 2$ back into the equation:
$y cos x = ln ∣ \frac{t a n ( \frac{x}{2} ) + 1}{t a n ( \frac{x}{2} ) + 2} ∣ + ln 2$
Combining logs:
$y cos x = ln ∣ \frac{2 ( t a n ( \frac{x}{2} ) + 1 )}{t a n ( \frac{x}{2} ) + 2} ∣$

Evaluating at $x = \frac{π}{3}$

At $x = \frac{π}{3}$ :
$cos (\frac{π}{3}) = \frac{1}{2}$
$tan (\frac{x}{2}) = tan (\frac{π}{6}) = \frac{1}{3}$
Substituting these values:
$\frac{1}{2} y = ln ∣ \frac{2 ( \frac{1}{3} + 1 )}{\frac{1}{3} + 2} ∣ = ln ∣ \frac{2 ( \frac{1 + 3}{3} )}{\frac{1 + 2 3}{3}} ∣$
$\frac{1}{2} y = ln ∣ \frac{2 3 + 2}{2 3 + 1} ∣$

Rationalizing the Argument

Rationalizing the fraction inside the log:
$\frac{2 3 + 2}{2 3 + 1} \cdot \frac{2 3 - 1}{2 3 - 1} = \frac{( 2 3 ) ( 2 3 ) - 2 3 + 4 3 - 2}{( 2 3 ) ^{2} - 1 ^{2}}$
Simplifying:
$\frac{12 + 2 3 - 2}{12 - 1} = \frac{10 + 2 3}{11}$
So, $\frac{1}{2} y = ln ∣ \frac{10 + 2 3}{11} ∣$

Final Result

Final Value of $y$ :
$y = 2 ln ∣ \frac{10 + 2 3}{11} ∣$
Key Takeaways:
1. Identified and solved a Linear Differential Equation.
2. Used the Integrating Factor method ( $I . F . = cos x$ ).
3. Applied half-angle substitution ( $t = tan (\frac{x}{2})$ ) for the integral.
4. Used initial conditions to find the particular solution.

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The given differential equation is:

cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x

To solve this, we first rearrange the equation into the standard form of a Linear Differential Equation (LDE):

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

Dividing both sides by

d x

and the term

cos x (3 sin x + cos x + 3)

, we obtain:

\frac{d y}{d x} = \frac{1 + y sin x ( 3 sin x + cos x + 3 )}{cos x ( 3 sin x + cos x + 3 )}

The Master Equation

We split the fraction on the right-hand side into two distinct parts:

\frac{d y}{d x} = \frac{1}{cos x ( 3 sin x + cos x + 3 )} + \frac{y sin x ( 3 sin x + cos x + 3 )}{cos x ( 3 sin x + cos x + 3 )}

The term

(3 sin x + cos x + 3)

cancels out in the second part, leaving us with

y tan x

. Moving this term to the left side yields the standard LDE form:

\frac{d y}{d x} - (tan x) y = \frac{1}{cos x ( 3 sin x + cos x + 3 )}

The Integrating Factor

To solve this LDE, we calculate the Integrating Factor (I.F.):

I . F . = e^{\int P (x) d x} = e^{\int - t a n x d x} = e^{l n ∣ c o s x ∣} = cos x

Multiplying the entire LDE by

cos x

, the left side becomes the derivative of the product

y cos x

. The right side simplifies significantly:

\frac{d}{d x} (y cos x) = \frac{1}{3 sin x + cos x + 3}

Integrating both sides, we must solve:

y cos x = \int \frac{d x}{3 sin x + cos x + 3}

The Weierstrass Transformation

We apply the half-angle substitution

t = tan (\frac{x}{2})

, where

sin x = \frac{2 t}{1 + t ^{2}}

cos x = \frac{1 - t ^{2}}{1 + t ^{2}}

, and

d x = \frac{2 d t}{1 + t ^{2}}

y cos x = \int \frac{\frac{2 d t}{1 + t ^{2}}}{3 ( \frac{2 t}{1 + t ^{2}} ) + ( \frac{1 - t ^{2}}{1 + t ^{2}} ) + 3}

Simplifying the denominator leads to:

y cos x = \int \frac{2 d t}{2 t ^{2} + 6 t + 4} = \int \frac{d t}{t ^{2} + 3 t + 2}

Using partial fractions, we resolve the integral:

y cos x = \int (\frac{1}{t + 1} - \frac{1}{t + 2}) d t = ln \frac{t + 1}{t + 2} + C

Final Calculation

Substituting

t = tan (\frac{x}{2})

back into the equation, the general solution is:

y cos x = ln \frac{tan ( \frac{x}{2} ) + 1}{tan ( \frac{x}{2} ) + 2} + C

Using the initial condition

y (0) = 0

, we find

0 = ln (1/2) + C

, which implies

C = ln 2

. Thus, the particular solution is:

y cos x = ln \frac{2 ( tan ( \frac{x}{2} ) + 1 )}{tan ( \frac{x}{2} ) + 2}

Evaluating at

x = \frac{π}{3}

, where

cos (\frac{π}{3}) = \frac{1}{2}

and

tan (\frac{π}{6}) = \frac{1}{3}

, we arrive at the final result: $y = 2 ln \frac{10 + 2 3}{11}$

Common Traps & FAQs

Why can we just divide by the cosine term? Isn't that mathematically dangerous?

Great question! In the context of JEE problems, we assume the domain is restricted where the denominator is non-zero. Here,

0 \leq x \leq \frac{π}{2}

, so

cos x

is positive and the expression

3 sin x + cos x + 3

is strictly positive. It is perfectly safe to divide!

I got stuck on the integral

\int \frac{d x}{3 s i n x + c o s x + 3}

. Is there a faster way than half-angle substitution?

While you could try harmonic addition (the R-method), the half-angle substitution is the most robust, 'fail-safe' method for rational functions of sine and cosine. It turns a tricky trig integral into a simple partial fraction problem. Trust the process!

Why did we need the constant

C

Differential equations represent a family of curves. The constant

C

represents the 'vertical shift' of these curves. The initial condition

y (0) = 0

acts like a GPS coordinate, pinning down exactly which curve in that family is our solution.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Differential Equation

Rearranging for $\frac{d y}{d x}$

Identifying the Linear Form

Calculating the Integrating Factor (I.F.)

Setting Up the General Solution

Half-Angle Substitution Method

Transforming the Integral to $t$

Simplifying the Quadratic Expression

Integrating via Partial Fractions

Applying the Initial Condition

The Particular Solution

Evaluating at $x = \frac{π}{3}$

Rationalizing the Argument

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

The Integrating Factor

The Weierstrass Transformation

Final Calculation

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $cos x (lo g_{e} (cos x))^{2} d y + (sin x - 3 y sin x lo g_{e} (cos x)) d x = 0$ $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{l o g _{e} 2}$ , then $y (\frac{π}{6})$ 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:

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to

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

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ 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

Let the solution curve $y = y (x)$ of the differential equation $\frac{d y}{d x} + \frac{3 x ^{5} t a n ^{- 1} ( x ^{3} )}{( 1 + x ^{6} ) ^{\frac{3}{2}}} y = 2 x exp (\frac{x ^{3} - t a n ^{- 1} x ^{3}}{( 1 + x ^{6} )})$ pass through the origin. Then $y (1)$ 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 :

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

Select Answer:

Visualized Solution

Analyze the Given Differential Equation

Rearranging for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Identifying the Linear Form

Calculating the Integrating Factor (I.F.)

Setting Up the General Solution

Half-Angle Substitution Method

Transforming the Integral to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Simplifying the Quadratic Expression

Integrating via Partial Fractions

Applying the Initial Condition

The Particular Solution

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3π​

Rationalizing the Argument

Final Result

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

The Integrating Factor

The Weierstrass Transformation

Final Calculation

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 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; } If y=y(x) is the solution curve of the differential equation dxdy​+ytanx=xsecx,0≤x≤3π​,y(0)=1, then y(6π​) 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​+3(tan2x)y+3y=sec2x, y(0)=31​+e3. Then y(4π​) 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 curve of the differential equation dxdy​=xy​(1+xy2(1+loge​x)),x>0,y(1)=3. Then 9y2(x)​ 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​=sinx(secx−sinx tanx)(tanx)+y​,x∈(0,2π​) satisfying the condition y(4π​)=2. Then, y(3π​) is

.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; } If y=y(x) is the solution of the equation esinycosydxdy​+esinycosx=cosx,y(0)=0; then 1+y(2π​)+2π​​y′(2π​)+2​1​y′′(2π​) 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 secxdxdy​−2y=2+3sinx,x∈(−2π​,2π​),y(0)=−47​. Then y(6π​) is equal to :

Rearranging for $\frac{d y}{d x}$

Transforming the Integral to $t$

Evaluating at $x = \frac{π}{3}$

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:

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to

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

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ 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 $y = y (x)$ 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 :

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