JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If a curve $y = y (x)$ passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} cot y - e^{x} csc y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at $x = 2$ , the value of $cos y$ is:

Q: How did you know to multiply by $\sin y$?

This is a classic pattern recognition skill. When you see $\cot y$ and $\csc y$ in a differential equation, they are essentially $\cos y/\sin y$ and $1/\sin y$. Multiplying by $\sin y$ is a strategic move to clear the denominators and transform the trigonometric functions into a form that matches the derivative of a function (in this case, $\cos y$). It is all about simplifying the expression to reveal the underlying linear structure.

Q: Why did we choose $t = -\cos y$ instead of just $t = \cos y$?

It is all about the signs! If we let $t = \cos y$, then $dt/dx = -\sin y (dy/dx)$. If we let $t = -\cos y$, then $dt/dx = \sin y (dy/dx)$. Looking at our equation, we had a positive $\sin y (dy/dx)$ term. Choosing $t = -\cos y$ makes the substitution cleaner and avoids unnecessary negative signs during the integration process. It is a small choice that prevents big mistakes!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look at the structure of the differential equation. It is currently expressed as a function of

d x / d y

. This is not the standard form we are used to. Try to invert the equation to express

d y / d x

in terms of

x

and

y

. Once you do that, you will see a structure that hints at a potential linear differential equation, but it is currently obscured by trigonometric functions.

Hint 2

Hint 2 (The Arsenal): Once you have

d y / d x

, you will notice terms involving

cot y

and

csc y

. Recall that

csc y = 1/ sin y

and

cot y = cos y / sin y

. If you multiply the entire equation by

sin y

, you can eliminate the cosecant term and simplify the cotangent term into a cosine term. This will set the stage for a substitution that linearizes the equation.

Hint 3

Hint 3 (The Execution): After multiplying by

sin y

, you should have an equation involving

sin y (d y / d x)

and

cos y

. Let

t = - cos y

. Then,

d t / d x = sin y (d y / d x)

. Substitute these into your equation to obtain a linear differential equation of the form

d t / d x + P (x) t = Q (x)

. From here, you can find the integrating factor

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

and solve for

t

Visualized Solution

Analyze the Differential Equation

Given equation: $(7 x^{4} cot y - e^{x} csc y) \frac{d x}{d y} = x^{5}$
Initial condition: Curve passes through $(1, \frac{π}{2})$
Goal: Find $cos y$ at $x = 2$

Rearranging the Equation

Rearranging to find $\frac{d y}{d x}$ :
$\frac{d y}{d x} = \frac{7 x ^{4} c o t y - e ^{x} c s c y}{x ^{5}}$

Simplify the Expression

Divide by $x^{5}$ : $\frac{d y}{d x} = \frac{7 c o t y}{x} - \frac{e ^{x} c s c y}{x ^{5}}$
Multiply by $sin y$ :
$sin y \frac{d y}{d x} - \frac{7 c o s y}{x} = - \frac{e ^{x}}{x ^{5}}$

Apply Substitution

Let $t = - cos y$
Differentiating with respect to $x$ :
$\frac{d t}{d x} = sin y \frac{d y}{d x}$

Form the Linear Differential Equation

Substitute $t$ and $\frac{d t}{d x}$ into the equation:
$\frac{d t}{d x} + \frac{7}{x} t = - \frac{e ^{x}}{x ^{5}}$
This is a linear differential equation of the form $\frac{d t}{d x} + P (x) t = Q (x)$

Calculate the Integrating Factor (IF)

$I F = e^{\int P (x) d x} = e^{\int \frac{7}{x} d x}$
$I F = e^{7 l n x} = e^{l n x^{7}}$
$I F = x^{7}$

Solve the Linear Equation

Solution format: $t \cdot (I F) = \int Q (x) \cdot (I F) d x$
$t \cdot x^{7} = \int - \frac{e ^{x}}{x ^{5}} \cdot x^{7} d x$
$t \cdot x^{7} = - \int x^{2} e^{x} d x$

Integration by Parts

Using Integration by Parts: $\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$
$\int x^{2} e^{x} d x = x^{2} e^{x} - 2 x e^{x} + 2 e^{x}$
So, $t \cdot x^{7} = - (x^{2} e^{x} - 2 x e^{x} + 2 e^{x}) + C$

Substitute back and find Constant C

Substitute $t = - cos y$ : $- cos y \cdot x^{7} = - e^{x} (x^{2} - 2 x + 2) + C$
At $(1, \frac{π}{2})$ : $- cos (\frac{π}{2}) \cdot 1^{7} = - e^{1} (1^{2} - 2 (1) + 2) + C$
$0 = - e (1) + C \Rightarrow C = e$

Final Calculation at $x = 2$

General equation: $cos y \cdot x^{7} = e^{x} (x^{2} - 2 x + 2) - e$
At $x = 2$ : $cos y \cdot 2^{7} = e^{2} (2^{2} - 2 (2) + 2) - e$
$cos y \cdot 128 = 2 e^{2} - e$
$cos y = \frac{2 e ^{2} - e}{128}$

Conclusion and Key Takeaway

Key Takeaway:
Transforming non-linear equations into linear ones using substitution is a powerful technique.
Always check for integrating factors in first-order equations.
Final Value: $cos y = \frac{2 e ^{2} - e}{128}$

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

We are given the differential equation

(7 x^{4} cot y - e^{x} csc y) \frac{d x}{d y} = x^{5}

. The presence of

\frac{d x}{d y}

suggests we should invert the equation to work with the more familiar

\frac{d y}{d x}

By taking the reciprocal, we transform the equation into:

\frac{d y}{d x} = \frac{7 x ^{4} cot y - e ^{x} csc y}{x ^{5}}

Distributing the

x^{5}

in the denominator yields:

\frac{d y}{d x} = \frac{7 cot y}{x} - \frac{e ^{x} csc y}{x ^{5}}

The Magic of Sine

The trigonometric terms

cot y

and

csc y

are the primary obstacles. Recalling the identities

csc y = \frac{1}{s i n y}

and

cot y = \frac{c o s y}{s i n y}

, we multiply the entire equation by

sin y

sin y \frac{d y}{d x} = \frac{7 cos y}{x} - \frac{e ^{x}}{x ^{5}}

Rearranging the terms to group the

y

-dependent functions on the left, we obtain:

sin y \frac{d y}{d x} - \frac{7 cos y}{x} = - \frac{e ^{x}}{x ^{5}}

The Linear Path

We observe that the derivative of

cos y

involves

sin y

, which suggests a substitution. Let

t = - cos y

Differentiating with respect to

x

, we get

\frac{d t}{d x} = sin y \frac{d y}{d x}

. Substituting this into our equation, we arrive at:

\frac{d t}{d x} + \frac{7}{x} t = - \frac{e ^{x}}{x ^{5}}

This is now a standard first-order Linear Differential Equation of the form

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

The Integrating Factor

To solve this, we calculate the Integrating Factor (

I F

I F = e^{\int P (x) d x} = e^{\int \frac{7}{x} d x} = e^{7 l n x} = x^{7}

Multiplying the linear equation by

x^{7}

, we get:

t \cdot x^{7} = \int - \frac{e ^{x}}{x ^{5}} \cdot x^{7} d x = - \int x^{2} e^{x} d x

The Final Stretch

Performing integration by parts on

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

twice, we obtain

x^{2} e^{x} - 2 x e^{x} + 2 e^{x}

. Thus, the general solution is:

t \cdot x^{7} = - (x^{2} e^{x} - 2 x e^{x} + 2 e^{x}) + C

Substituting

t = - cos y

back into the equation:

cos y \cdot x^{7} = e^{x} (x^{2} - 2 x + 2) - C

Using the initial condition

(1, \frac{π}{2})

, where

cos (\frac{π}{2}) = 0

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

Finally, evaluating at

x = 2

cos y \cdot 2^{7} = e^{2} (2^{2} - 2 (2) + 2) - e

cos y \cdot 128 = 2 e^{2} - e

The final result is:

cos y = \frac{2 e ^{2} - e}{128}

Common Traps & FAQs

Wait, why do we flip the equation from

d x / d y

d y / d x

Great question! In JEE problems, we are almost always dealing with

y

as a function of

x

. The given equation was presented in a slightly disguised form to test your flexibility. By flipping it to

d y / d x

, we align the problem with the standard linear differential equation form

d y / d x + P (x) y = Q (x)

, which allows us to use the powerful Integrating Factor method.

How did you know to multiply by

sin y

This is a classic pattern recognition skill. When you see

cot y

and

csc y

in a differential equation, they are essentially

cos y / sin y

and

1/ sin y

. Multiplying by

sin y

is a strategic move to clear the denominators and transform the trigonometric functions into a form that matches the derivative of a function (in this case,

cos y

). It is all about simplifying the expression to reveal the underlying linear structure.

Why did we choose

t = - cos y

instead of just

t = cos y

It is all about the signs! If we let

t = cos y

, then

d t / d x = - sin y (d y / d x)

. If we let

t = - cos y

, then

d t / d x = sin y (d y / d x)

. Looking at our equation, we had a positive

sin y (d y / d x)

term. Choosing

t = - cos y

makes the substitution cleaner and avoids unnecessary negative signs during the integration process. It is a small choice that prevents big mistakes!

Animated Solution for Mathematics - Differential Equations: If a curve $y = y (x)$ passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} cot y - e^{x} csc y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at $x = 2$ , the value of $cos y$ is:

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging the Equation

Simplify the Expression

Apply Substitution

Form the Linear Differential Equation

Calculate the Integrating Factor (IF)

Solve the Linear Equation

Integration by Parts

Substitute back and find Constant C

Final Calculation at $x = 2$

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of Sine

The Linear Path

The Integrating Factor

The Final Stretch

Similar Questions

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:

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 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is 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

If the curve $y = y (x)$ is the solution of the differential equation $2 (x^{2} + x^{5/4}) d y - y (x + x^{1/4}) d x = 2 x^{9/4} d x, x > 0$ which passes through the point $(1, 1 - \frac{4}{3} lo g_{e} 2)$ , then the value of $y (16)$ 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 $cos x (lo g_{e} (cos x))^{2} d y + (sin x - 3 y sin x lo g_{e} (cos x)) d x = 0$ $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{4}) = \frac{- 1}{l o g _{e} 2}$ , then $y (\frac{π}{6})$ is equal to :

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Rearranging the Equation

Simplify the Expression

Apply Substitution

Form the Linear Differential Equation

Calculate the Integrating Factor (IF)

Solve the Linear Equation

Integration by Parts

Substitute back and find Constant C

Final Calculation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of Sine

The Linear Path

The Integrating Factor

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 cosec2xdy+2dx=(1+ycos2x)cosec2xdx, with y(4π​)=0. Then, the value of (y(0)+1)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 dxdy​=2(y+2sinx−5)x−2cosx such that y(0)=7. Then y(π) 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​=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy​ at x=1 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 xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(e) 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 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 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 y=y(x) is the solution of the differential equation xdxdy​+2y=x2 satisfying y(1)=1, then y(1/2) is equal to :

Final Calculation at $x = 2$

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:

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 :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is 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 $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :