JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

Q: How do I know when to use the constant of integration $C$?

Always include $C$ immediately after performing the indefinite integration. It represents the family of all possible curves that satisfy the given slope condition. The boundary condition (the point $(2, (\ln 2)^2)$) acts as a 'filter' that selects the one specific curve from that family that passes through our target point.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the curve on the Cartesian plane. The problem gives you the slope of the tangent at any point

(x, y)

. Recall that the slope of a curve

y = f (x)

is defined as the derivative

\frac{d y}{d x}

. This is your gateway into the world of differential equations.

Hint 2

Hint 2: The Arsenal. You have a first-order differential equation:

\frac{d y}{d x} = \frac{2 y}{x l n x}

. To solve this, you must separate the variables. Group all terms involving

y

on one side and all terms involving

x

on the other. Once separated, the integration of the right-hand side will require a simple substitution: let

u = ln x

Hint 3

Hint 3: The Execution. After separating the variables, you will have

\int \frac{1}{y} d y = \int \frac{2}{x l n x} d x

. Integrate both sides to get

ln y = 2 ln (ln x) + C

. Use the given point

(2, (ln 2)^{2})

to determine the constant

C

before evaluating the function at

x = e

Visualized Solution

Visualizing the Problem

Given curve: $y = f (x)$
Slope at any point: $\frac{2 y}{x l n x}$

Defining the Differential Equation

The slope of a curve is given by its derivative, $\frac{d y}{d x}$ .
Therefore, $\frac{d y}{d x} = \frac{2 y}{x l n x}$

Separating the Variables

Rearrange terms to group $y$ with $d y$ and $x$ with $d x$ .
$\frac{1}{y} d y = \frac{2}{x l n x} d x$

Integration Strategy

Integrate both sides: $\int \frac{1}{y} d y = \int \frac{2}{x l n x} d x$
For the RHS, use substitution: Let $u = ln x$ , then $d u = \frac{1}{x} d x$ .

Executing the Integration

LHS: $\int \frac{1}{y} d y = ln y$
RHS: $2 \int \frac{1}{u} d u = 2 ln u + C$
Substitute back $u = ln x$ : $ln y = 2 ln (ln x) + C$

Simplifying the General Solution

Use the logarithmic property: $n ln a = ln (a^{n})$
$ln y = ln ((ln x)^{2}) + C$

Applying the Boundary Condition

The curve passes through the point $(2, (ln 2)^{2})$ .
Substitute $x = 2$ and $y = (ln 2)^{2}$ into the general solution.
$ln ((ln 2)^{2}) = ln ((ln 2)^{2}) + C$

Finding the Constant $C$

$ln ((ln 2)^{2}) = ln ((ln 2)^{2}) + C$
Subtracting $ln ((ln 2)^{2})$ from both sides gives:
$C = 0$

The Particular Solution

Substitute $C = 0$ back into the simplified general solution:
$ln y = ln ((ln x)^{2})$
Taking the antilog (exponential) on both sides:
$y = (ln x)^{2}$

Evaluating $f (e)$

We need to find the value of $f (e)$ .
Substitute $x = e$ into $f (x) = (ln x)^{2}$ .
$f (e) = (ln e)^{2}$

Final Conclusion

Recall that the natural logarithm of $e$ is $1$ , so $ln e = 1$ .
$f (e) = (1)^{2} = 1$
The final answer is $1$ .

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The slope of the tangent at any point

(x, y)

on the curve

y = f (x)

is defined by the expression:

\frac{d y}{d x} = \frac{2 y}{x ln x}

This differential equation describes the growth rate of the function. Our objective is to determine the specific function

f (x)

and evaluate it at

x = e

The Art of Separation

To solve this, we employ the method of separation of variables. We group all terms involving

y

on the left and all terms involving

x

on the right:

\frac{1}{y} d y = \frac{2}{x ln x} d x

Integrating both sides of the equation yields:

\int \frac{1}{y} d y = \int \frac{2}{x ln x} d x

The Integration Dance

The left side integrates directly to

ln y

. For the right side, we use the substitution method where

u = ln x

, which implies

d u = \frac{1}{x} d x

Substituting these into the integral, we obtain:

\int \frac{1}{y} d y = 2 \int \frac{1}{u} d u

ln y = 2 ln u + C

Substituting

u = ln x

back into the equation, we arrive at the general solution:

ln y = 2 ln (ln x) + C

The Boundary Condition

We are given that the curve passes through the point

(2, (ln 2)^{2})

. Substituting

x = 2

and

y = (ln 2)^{2}

into our general solution allows us to solve for the constant

C

ln ((ln 2)^{2}) = 2 ln (ln 2) + C

Using the logarithmic property

n ln a = ln (a^{n})

, we observe that

2 ln (ln 2) = ln ((ln 2)^{2})

. This simplifies the equation to:

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

The Final Reveal

With

C = 0

, the equation simplifies to

ln y = 2 ln (ln x)

, which is equivalent to

ln y = ln ((ln x)^{2})

. Exponentiating both sides, we find the function:

y = (ln x)^{2}

To find the final value at

x = e

, we substitute into the function:

f (e) = (ln e)^{2}

Since

ln e = 1

, we calculate:

$f (e) = 1^{2} = 1$

Common Traps & FAQs

Wait, why do we reject the negative root here? Or do we?

Great question! In this specific problem, we are dealing with the natural logarithm

ln x

, which is only defined for

x > 0

. When we integrate

\frac{1}{y} d y

, we get

ln ∣ y ∣

. However, given the point

(2, (ln 2)^{2})

, we know

y

must be positive, so we can safely work with

ln y

. The structure of the equation

y = (ln x)^{2}

naturally keeps

y

positive for

x > 1

Why can't we just use energy conservation or some other physics principle?

This is a pure calculus problem disguised as a curve-tracing exercise! While physics often uses differential equations (like in simple harmonic motion), here the 'physics' is simply the geometric definition of a slope. We are not dealing with forces or energy, but with the rate of change of a function. Stick to the calculus toolkit!

How do I know when to use the constant of integration

C

Always include

C

immediately after performing the indefinite integration. It represents the family of all possible curves that satisfy the given slope condition. The boundary condition (the point

(2, (ln 2)^{2})

) acts as a 'filter' that selects the one specific curve from that family that passes through our target point.

Animated Solution for Mathematics - Differential Equations: Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Defining the Differential Equation

Separating the Variables

Integration Strategy

Executing the Integration

Simplifying the General Solution

Applying the Boundary Condition

Finding the Constant $C$

The Particular Solution

Evaluating $f (e)$

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Dance

The Boundary Condition

The Final Reveal

Similar Questions

Let a smooth curve $y = f (x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\frac{- y}{x}$ . If the curve passes through the point $(1, 2)$ and $(8, 1)$ , then $∣ y (\frac{1}{8}) ∣$ 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 $f (x) = lim_{r \to x} {\frac{2 r ^{2} [( f ( r ) ) ^{2} - f ( x ) f ( r )]}{r ^{2} - x ^{2}} - r^{3} e^{\frac{f ( r )}{r}}}$ be differentiable in $(- \infty, 0) \cup (0, \infty)$ and $f (1) = 1$ . Then the value of $e a$ , such that $f (a) = 0$ , is equal to ______.

Let $f : [0, 2] \to R$ be a function which is continuous on $[0, 2]$ and is differentiable on $(0, 2)$ with $f (0) = 1$ . Let $F (x) = \int_{0}^{x^{2}} f (t) d t$ for $x \in [0, 2]$ . If $F^{'} (x) = f^{'} (x)$ for all $x \in (0, 2)$ , then $f (2)$ equals

If $y (x)$ is the solution of the differential equation $x d y - (y^{2} - 4 y) d x = 0$ for $x > 0$ , $y (1) = 2$ , and the slope of the curve $y = y (x)$ is never zero, then the value of $10 y (2)$ is ______.

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Given that the slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is $2 y / x^{2}$ . If the curve passes through the centre of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$ , then its equation is :

Let a curve $y = y (x)$ be given by the solution of the differential equation $cos (\frac{1}{2} cos^{- 1} (e^{- x})) d x = e^{2 x} - 1 d y$ . If it intersects $y$ -axis at $y = - 1$ , and the intersection point of the curve with $x$ -axis is $(α, 0)$ , then $e^{α}$ is equal to

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k 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

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Defining the Differential Equation

Separating the Variables

Integration Strategy

Executing the Integration

Simplifying the General Solution

Applying the Boundary Condition

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Particular Solution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(e)

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Dance

The Boundary Condition

The Final Reveal

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If length of tangent at any point on the curve y=f(x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a curve y=f(x) pass through the points (0,5) and (loge​2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=0, then k 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

Finding the Constant $C$

Evaluating $f (e)$

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

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k 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