JEE Main 2024 (08 April Shift 2)

LEVELJEE Main

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

Q: Why can't we just separate the variables directly?

Because the $x$ and $y$ terms are inextricably linked through addition. Separation of variables only works if you can write the equation as $f(y)dy = g(x)dx$. Here, the structure demands a linear transformation first to isolate the variables effectively.

Q: How did you know to divide by $\cos y$?

It is all about pattern recognition. Seeing $\sec y$ and $\sin y$ together screams 'tangent' and 'secant squared'. By dividing by $\cos y$, we force the equation into a form where the derivative of the function ($\tan y$) is present, which is the hallmark of a solvable differential equation.

Q: Is the integration by parts step necessary?

Absolutely. The integral $\int x^3 e^{x^2} dx$ cannot be solved by simple inspection. By substituting $u = x^2$, we reduce it to $\frac{1}{2} \int u e^u du$, which is a classic integration by parts problem. It is a beautiful example of how substitution and parts work together.

Select Answer:

Visualized Solution

Identify the Differential Equation

Given equation: $sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y$
Initial condition: $y (1) = 0$
Goal: Find $y (3)$

Simplify by Dividing by $cos y$

Divide the entire equation by $cos y$ :
$\frac{s e c y}{c o s y} \frac{d y}{d x} + 2 x \frac{s i n y}{c o s y} = x^{3}$
Using $\frac{1}{c o s y} = sec y$ and $\frac{s i n y}{c o s y} = tan y$ :
$sec^{2} y \frac{d y}{d x} + 2 x tan y = x^{3}$

Substitution to Linear Form

Let $t = tan y$
Differentiating both sides with respect to $x$ :
$\frac{d t}{d x} = sec^{2} y \frac{d y}{d x}$

Form the Linear Differential Equation

Substitute $t$ and $\frac{d t}{d x}$ into the simplified equation:
$\frac{d t}{d x} + 2 x t = x^{3}$
This is a standard Linear Differential Equation (LDE) of the form $\frac{d t}{d x} + P (x) t = Q (x)$
Here, $P (x) = 2 x$ and $Q (x) = x^{3}$

Calculate the Integrating Factor

Integrating Factor $I F = e^{\int P (x) d x}$
Substitute $P (x) = 2 x$ :
$I F = e^{\int 2 x d x}$
$I F = e^{x^{2}}$

Write the General Solution

The general solution for an LDE is: $t \cdot (I F) = \int Q (x) \cdot (I F) d x + C$
Substitute $t$ , $I F$ , and $Q (x)$ :
$tan y \cdot e^{x^{2}} = \int x^{3} e^{x^{2}} d x + C$

Solve the Integral: Substitution

Let's evaluate $I = \int x^{3} e^{x^{2}} d x$
Rewrite $x^{3}$ as $x^{2} \cdot x$ :
$I = \int x^{2} e^{x^{2}} x d x$
Let $u = x^{2}$ , then $d u = 2 x d x$ , which means $x d x = \frac{d u}{2}$
$I = \int u e^{u} \frac{d u}{2}$

Solve the Integral: By Parts

$I = \frac{1}{2} \int u e^{u} d u$
Using Integration by Parts: $\int u e^{u} d u = u e^{u} - \int e^{u} d u = u e^{u} - e^{u}$
$I = \frac{1}{2} (u e^{u} - e^{u}) = \frac{1}{2} e^{u} (u - 1)$
Substitute back $u = x^{2}$ :
$I = \frac{1}{2} e^{x^{2}} (x^{2} - 1)$

Complete the General Solution

Substitute the integral back into our equation:
$tan y \cdot e^{x^{2}} = \frac{1}{2} e^{x^{2}} (x^{2} - 1) + C$
Divide the entire equation by $e^{x^{2}}$ :
$tan y = \frac{1}{2} (x^{2} - 1) + C e^{- x^{2}}$

Apply Initial Condition

We are given $y (1) = 0$
This means when $x = 1$ , $y = 0$
Substitute these values into the general solution:
$tan (0) = \frac{1}{2} (1^{2} - 1) + C e^{- 1^{2}}$

Find the Constant of Integration

Evaluate the terms: $tan (0) = 0$ and $1^{2} - 1 = 0$
$0 = \frac{1}{2} (0) + C e^{- 1}$
$0 = C e^{- 1}$
Since $e^{- 1}$ is not zero, $C$ must be $0$ .
The specific solution is: $tan y = \frac{x ^{2} - 1}{2}$

Calculate $y (3)$

We need to find $y$ when $x = 3$
Substitute $x = 3$ into the specific solution:
$tan y = \frac{( 3 ) ^{2} - 1}{2}$
$tan y = \frac{3 - 1}{2} = \frac{2}{2} = 1$
$y = tan^{- 1} (1) = \frac{π}{4}$

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The Sigma Insight: Linear Differential Equations

Analyzing the Setup

When you first look at the equation

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

, it is natural to feel a moment of hesitation. In the world of JEE Advanced, complexity is often just a disguise for elegance. Our mission is to strip away that disguise.

Phase 1

The Simplification

The first step is to observe the structure. We have

sec y

sin y

, and

cos y

all mixed with

x

. The key insight here is to simplify the trigonometric landscape.

By dividing the entire equation by

cos y

, we perform a surgical strike on the complexity. The term

sec y

becomes

sec^{2} y

, and

sin y

becomes

tan y

. Suddenly, we are looking at:

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

This is not just a cleaner equation; it is a revelation.

Phase 2

The Transformation

Now, look at the term

sec^{2} y \frac{d y}{d x}

. It is the derivative of

tan y

with respect to

x

. This is the "Aha!" moment.

We introduce a substitution: let

t = tan y

. Consequently,

\frac{d t}{d x} = sec^{2} y \frac{d y}{d x}

. Our equation transforms into:

\frac{d t}{d x} + 2 x t = x^{3}

We have successfully mapped a non-linear nightmare into a standard Linear Differential Equation (LDE) of the form

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

Phase 3

The Linear Engine

With

P (x) = 2 x

and

Q (x) = x^{3}

, we are on familiar ground. We calculate the Integrating Factor (

I F

I F = e^{\int 2 x d x} = e^{x^{2}}

This factor is the magic key. Multiplying our LDE by

e^{x^{2}}

allows us to collapse the left side into the derivative of a product:

\frac{d}{d x} (t \cdot e^{x^{2}}) = x^{3} e^{x^{2}}

Phase 4

The Integration Challenge

Now, we face the integral

\int x^{3} e^{x^{2}} d x

. We use substitution again: let

u = x^{2}

, so

d u = 2 x d x

. The integral becomes:

\frac{1}{2} \int u e^{u} d u

This is a classic Integration by Parts scenario. Solving this yields

\frac{1}{2} e^{u} (u - 1)

, which translates back to:

\frac{1}{2} e^{x^{2}} (x^{2} - 1)

Conclusion

The Final Victory

We combine everything:

tan y \cdot e^{x^{2}} = \frac{1}{2} e^{x^{2}} (x^{2} - 1) + C

Applying the initial condition

y (1) = 0

, we find that

C = 0

. Our specific solution is:

tan y = \frac{x ^{2} - 1}{2}

Finally, substituting

x = 3

, we get

tan y = 1

, which means $y = \frac{π}{4}$ . You have navigated the storm and arrived at the solution.

Common Traps & FAQs

Why can't we just separate the variables directly?

Because the

x

and

y

terms are inextricably linked through addition. Separation of variables only works if you can write the equation as

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

. Here, the structure demands a linear transformation first to isolate the variables effectively.

How did you know to divide by

cos y

It is all about pattern recognition. Seeing

sec y

and

sin y

together screams 'tangent' and 'secant squared'. By dividing by

cos y

, we force the equation into a form where the derivative of the function (

tan y

) is present, which is the hallmark of a solvable differential equation.

Is the integration by parts step necessary?

Absolutely. The integral

\int x^{3} e^{x^{2}} d x

cannot be solved by simple inspection. By substituting

u = x^{2}

, we reduce it to

\frac{1}{2} \int u e^{u} d u

, which is a classic integration by parts problem. It is a beautiful example of how substitution and parts work together.

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

Select Answer:

Visualized Solution

Identify the Differential Equation

Simplify by Dividing by $cos y$

Substitution to Linear Form

Form the Linear Differential Equation

Calculate the Integrating Factor

Write the General Solution

Solve the Integral: Substitution

Solve the Integral: By Parts

Complete the General Solution

Apply Initial Condition

Find the Constant of Integration

Calculate $y (3)$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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

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 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 curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

If $y = y (x), x \in (0, \frac{π}{2})$ be the solution curve of the differential equation $(sin^{2} 2 x) \frac{d y}{d x} + (8 sin^{2} 2 x + 2 sin 4 x) y = 2 e^{- 4 x} (2 sin 2 x + cos 2 x)$ , with $y (\frac{π}{4}) = e^{- π}$ , then $y (\frac{π}{6})$ is equal to :

Select Answer:

Visualized Solution

Identify the Differential Equation

Simplify by Dividing by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosy

Substitution to Linear Form

Form the Linear Differential Equation

Calculate the Integrating Factor

Write the General Solution

Solve the Integral: Substitution

Solve the Integral: By Parts

Complete the General Solution

Apply Initial Condition

Find the Constant of Integration

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3​)

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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 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; } 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 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 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 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 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; } If x=f(y) is the solution of the differential equation (1+y2)+(x−2etan−1y)dxdy​=0, y∈(−2π​,2π​) with f(0)=1, then f(3​1​) 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 x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(21​) is equal to :

Simplify by Dividing by $cos y$

Calculate $y (3)$

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

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:

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :