JEE Main 2022 (28 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Q: Can we just use substitution like $v = x/y^2$ instead of this exact differential method?

Absolutely! You can substitute $v = x/y^2$, which implies $x = vy^2$ and $dx = v(2y dy) + y^2 dv$. It works perfectly, but it is algebraically heavier and prone to calculation errors. The exact differential method is the 'pro' way to save time and keep your work clean.

Q: Why is the integral of $e^u du$ just $e^u$? Is it really that simple?

Yes! Once you transform the equation into $2e^u du + \frac{1}{y} dy = 0$, where $u = \frac{x}{y^2}$, the integration becomes trivial. The complexity was all in the setup. Once the setup is correct, the calculus is the reward for your hard work!

Select Answer:

Progressive Hints

Hint 1

Look closely at the structure of the equation

2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0

. Notice that the exponential term

e^{x / y^{2}}

appears in two distinct parts. Instead of trying to separate variables immediately, try grouping the terms containing this exponential factor together. What does the remaining expression

y d x - 2 x d y

suggest to you?

Hint 2

Recall the quotient rule for differentiation:

d (\frac{u}{v}) = \frac{v d u - u d v}{v ^{2}}

. Specifically, look at the differential of the exponent

d (\frac{x}{y ^{2}})

. Calculate this derivative and compare it to the expression

y d x - 2 x d y

you found in the previous step. How can you manipulate the equation to make them match?

Hint 3

To force the equation into an exact differential form, divide the entire equation by

y^{3}

. This will transform the term

(y d x - 2 x d y)

into the numerator of

d (\frac{x}{y ^{2}})

, allowing you to integrate the equation directly as

2 e^{x / y^{2}} d (\frac{x}{y ^{2}}) + \frac{1}{y} d y = 0

Visualized Solution

The Differential Equation

Given: $2 y e^{\frac{x}{y ^{2}}} d x + (y^{2} - 4 x e^{\frac{x}{y ^{2}}}) d y = 0$
Initial condition: $x (1) = 0$
Goal: Find $x (e)$

Expanding the Equation

Expand the brackets: $2 y e^{\frac{x}{y ^{2}}} d x + y^{2} d y - 4 x e^{\frac{x}{y ^{2}}} d y = 0$
Group the exponential terms together: $(2 y e^{\frac{x}{y ^{2}}} d x - 4 x e^{\frac{x}{y ^{2}}} d y) + y^{2} d y = 0$

Factoring the Exponential

Factor out $2 e^{\frac{x}{y ^{2}}}$ from the grouped terms.
Result: $2 e^{\frac{x}{y ^{2}}} (y d x - 2 x d y) + y^{2} d y = 0$

Creating an Exact Differential

We need to convert $y d x - 2 x d y$ into an exact differential.
Divide the entire equation by $y^{3}$ .
$2 e^{\frac{x}{y ^{2}}} \frac{y d x - 2 x d y}{y ^{3}} + \frac{y ^{2}}{y ^{3}} d y = 0$

Simplifying the Terms

Simplify the second term: $\frac{y ^{2}}{y ^{3}} d y = \frac{1}{y} d y$
The equation becomes: $2 e^{\frac{x}{y ^{2}}} \frac{y d x - 2 x d y}{y ^{3}} + \frac{1}{y} d y = 0$

The Quotient Rule

Recall the quotient rule: $d (\frac{u}{v}) = \frac{v d u - u d v}{v ^{2}}$
Let's find the differential of the exponent: $d (\frac{x}{y ^{2}})$
$d (\frac{x}{y ^{2}}) = \frac{y ^{2} d x - x ( 2 y d y )}{( y ^{2} ) ^{2}}$

Matching the Differential

Simplify the numerator: $d (\frac{x}{y ^{2}}) = \frac{y ( y d x - 2 x d y )}{y ^{4}}$
Cancel $y$ from numerator and denominator: $d (\frac{x}{y ^{2}}) = \frac{y d x - 2 x d y}{y ^{3}}$

Substituting the Exact Differential

Replace $\frac{y d x - 2 x d y}{y ^{3}}$ with $d (\frac{x}{y ^{2}})$
The simplified equation: $2 e^{\frac{x}{y ^{2}}} d (\frac{x}{y ^{2}}) + \frac{1}{y} d y = 0$

Integrating the Equation

Integrate both sides: $\int 2 e^{\frac{x}{y ^{2}}} d (\frac{x}{y ^{2}}) + \int \frac{1}{y} d y = C$
The integral of $e^{u} d u$ is $e^{u}$ .
The integral of $\frac{1}{y} d y$ is $ln ∣ y ∣$ .
Result: $2 e^{\frac{x}{y ^{2}}} + ln ∣ y ∣ = C$

Applying Initial Conditions

Use the given initial condition: $x (1) = 0$
This means when $y = 1$ , $x = 0$ .
Substitute these values into the general solution: $2 e^{\frac{0}{1 ^{2}}} + ln ∣1∣ = C$

Finding the Constant $C$

Evaluate the terms: $e^{0} = 1$ and $ln (1) = 0$
$2 (1) + 0 = C$
Therefore, $C = 2$
Particular solution: $2 e^{\frac{x}{y ^{2}}} + ln ∣ y ∣ = 2$

Finding $x (e)$

We need to find the value of $x$ when $y = e$ .
Substitute $y = e$ into the particular solution: $2 e^{\frac{x}{e ^{2}}} + ln (e) = 2$

Solving for the Exponential

We know that $ln (e) = 1$ .
Substitute this value: $2 e^{\frac{x}{e ^{2}}} + 1 = 2$
Isolate the exponential term: $2 e^{\frac{x}{e ^{2}}} = 1$
$e^{\frac{x}{e ^{2}}} = \frac{1}{2}$

Taking the Natural Logarithm

Take the natural log ( $ln$ ) on both sides: $ln (e^{\frac{x}{e ^{2}}}) = ln (\frac{1}{2})$
Simplify the left side: $\frac{x}{e ^{2}} = ln (\frac{1}{2})$
Use the logarithm property $ln (\frac{1}{a}) = - ln (a)$ : $\frac{x}{e ^{2}} = - ln (2)$

Final Answer

Multiply by $e^{2}$ to solve for $x$ : $x = - e^{2} ln (2)$
This matches Option 4: $- e^{2} lo g_{e} (2)$
Key Takeaway: Grouping terms and recognizing exact differentials is a powerful technique for solving complex differential equations.

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The Sigma Insight: Variable Separable Method

Analyzing the Setup

The given differential equation is:

2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0

First, we expand the brackets to reveal the underlying structure:

2 y e^{x / y^{2}} d x + y^{2} d y - 4 x e^{x / y^{2}} d y = 0

Identifying the Pattern

Now, we group the terms containing the exponential factor

e^{x / y^{2}}

2 e^{x / y^{2}} (y d x - 2 x d y) + y^{2} d y = 0

Observe the term

(y d x - 2 x d y)

. This structure strongly suggests the application of the quotient rule for differentials. Specifically, consider the differential of the ratio

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

d (\frac{x}{y ^{2}}) = \frac{y ^{2} d x - x ( 2 y d y )}{y ^{4}} = \frac{y ( y d x - 2 x d y )}{y ^{4}} = \frac{y d x - 2 x d y}{y ^{3}}

The Master Equation

To utilize this identity, we divide the entire differential equation by

y^{3}

2 e^{x / y^{2}} (\frac{y d x - 2 x d y}{y ^{3}}) + \frac{1}{y} d y = 0

Substituting the differential identity, the equation simplifies beautifully:

2 e^{x / y^{2}} d (\frac{x}{y ^{2}}) + \frac{1}{y} d y = 0

Final Integration and Solution

Integrating both sides with respect to their variables, we obtain:

2 e^{x / y^{2}} + ln ∣ y ∣ = C

Using the initial condition

x (1) = 0

(where

y = 1

and

x = 0

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

Thus, the general solution is

2 e^{x / y^{2}} + ln ∣ y ∣ = 2

. To find

x

when

y = e

2 e^{x / e^{2}} + ln (e) = 2

2 e^{x / e^{2}} + 1 = 2 \Rightarrow 2 e^{x / e^{2}} = 1 \Rightarrow e^{x / e^{2}} = \frac{1}{2}

Taking the natural logarithm on both sides:

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

The final result is: $x = - e^{2} ln (2)$

Common Traps & FAQs

Why did we divide by

y^{3}

? How was I supposed to guess that?

Great question! In JEE Advanced, we often look for the 'differential of a function'. We saw

y d x - 2 x d y

and knew the quotient rule for

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

involves

y^{2} d x - 2 x y d y

. By dividing by

y^{3}

, we essentially 'manufactured' the denominator needed to complete the differential

d (\frac{x}{y ^{2}})

. It is not guessing; it is reverse-engineering the derivative!

Can we just use substitution like

v = x / y^{2}

instead of this exact differential method?

Absolutely! You can substitute

v = x / y^{2}

, which implies

x = v y^{2}

and

d x = v (2 y d y) + y^{2} d v

. It works perfectly, but it is algebraically heavier and prone to calculation errors. The exact differential method is the 'pro' way to save time and keep your work clean.

Why is the integral of

e^{u} d u

just

e^{u}

? Is it really that simple?

Yes! Once you transform the equation into

2 e^{u} d u + \frac{1}{y} d y = 0

, where

u = \frac{x}{y ^{2}}

, the integration becomes trivial. The complexity was all in the setup. Once the setup is correct, the calculus is the reward for your hard work!

Animated Solution for Mathematics - Differential Equations: Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

Select Answer:

Visualized Solution

The Differential Equation

Expanding the Equation

Factoring the Exponential

Creating an Exact Differential

Simplifying the Terms

The Quotient Rule

Matching the Differential

Substituting the Exact Differential

Integrating the Equation

Applying Initial Conditions

Finding the Constant $C$

Finding $x (e)$

Solving for the Exponential

Taking the Natural Logarithm

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Identifying the Pattern

The Master Equation

Final Integration and Solution

Similar Questions

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

The solution of the equation $\frac{d ^{2} y}{d x ^{2}} = e^{- 2 x}$

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Let $x = x (y)$ be the solution of the differential equation $y = (x - y \frac{d x}{d y}) sin (\frac{x}{y})$ $y > 0$ and $x (1) = \frac{π}{2}$ Then $cos (x (2))$ is equal to :

Select Answer:

Visualized Solution

The Differential Equation

Expanding the Equation

Factoring the Exponential

Creating an Exact Differential

Simplifying the Terms

The Quotient Rule

Matching the Differential

Substituting the Exact Differential

Integrating the Equation

Applying Initial Conditions

Finding the Constant .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; } x(e)

Solving for the Exponential

Taking the Natural Logarithm

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Identifying the Pattern

The Master Equation

Final Integration and Solution

Similar Questions

.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, ey(dxdy​−1)=ex such that y(0)=0, then y(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 of the differential equation (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=(x−y)2, when y(1)=1, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x3dy+xydx=x2dy+2ydx; y(2)=e and x>1, then y(4) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=e. Then limx→0+​f(x) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solution of the differential equation ydx+(x+x2y)dy=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the equation dx2d2y​=e−2x

.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+y2)etanxdx+cos2x(1+e2tanx)dy=0,y(0)=1. 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 x=x(y) be the solution of the differential equation y=(x−ydydx​)sin(yx​) y>0 and x(1)=2π​ Then cos(x(2)) is equal to :

Finding the Constant $C$

Finding $x (e)$

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let $y = f (x)$ be the solution of the differential equation $y (x + 1) d x - x^{2} d y = 0, y (1) = e$ . Then $lim_{x \to 0^{+}} f (x)$ is equal to

Solution of the differential equation $y d x + (x + x^{2} y) d y = 0$ is

The solution of the equation $\frac{d ^{2} y}{d x ^{2}} = e^{- 2 x}$

Let $y = y (x)$ be the solution of the differential equation $(1 + y^{2}) e^{t a n x} d x + cos^{2} x (1 + e^{2 t a n x}) d y = 0, y (0) = 1$ . Then $y (\frac{π}{4})$ is equal to

Let $x = x (y)$ be the solution of the differential equation $y = (x - y \frac{d x}{d y}) sin (\frac{x}{y})$ $y > 0$ and $x (1) = \frac{π}{2}$ Then $cos (x (2))$ is equal to :