Let $y = y(x)$ be the solution of the differential equation $xdy - ydx = \sqrt{x^2 - y^2}dx, x \ge 1$, with $y(1) = 0$. If the area bounded by the line $x = 1, x = e^\pi, y = 0$ and $y = y(x)$ is $\alpha e^{2\pi} + \beta$, then the value of $10(\alpha + \beta)$ is equal to ____.

### Analyzing the Setup We begin with the differential equation: $$x dy - y dx = \sqrt{x^2 - y^2} dx$$ At first glance, this appears complex. However, we recognize that the term $x dy - y dx$ is the numerator of the derivative of the quotient $\frac{y}{x}$. By dividing both sides of the equation by $x^2$, we transform the expression into a recognizable form: $$d\left(\frac{y}{x}\right) = \frac{\sqrt{1 - (y/x)^2}}{x} dx$$ ### The Elegance of Substitution By separating the variables, we arrive at the following integral: $$\int \frac{d(y/x)}{\sqrt{1 - (y/x)^2}} = \int \frac{dx}{x}$$ The left side is the standard integral for $\sin^{-1}\left(\frac{y}{x}\right)$, while the right side yields $\ln x + C$. Applying the initial condition $y(1) = 0$, we find that $C = 0$. This reveals the hidden curve: $$y = x \sin(\ln x)$$ ### The Journey Through the Integral We must find the area $A$ bounded by this curve from $x = 1$ to $x = e^\pi$: $$A = \int_1^{e^\pi} x \sin(\ln x) dx$$ To solve this, we use the substitution $t = \ln x$, which implies $x = e^t$ and $dx = e^t dt$. The limits of integration change from $[1, e^\pi]$ to $[0, \pi]$. The integral becomes: $$A = \int_0^\pi e^{2t} \sin t dt$$ ### Evaluating the Integral We utilize the standard integration formula: $$\int e^{at} \sin bt dt = \frac{e^{at}}{a^2+b^2} (a \sin bt - b \cos bt)$$ Applying this with $a=2$ and $b=1$: $$A = \left[ \frac{e^{2t}}{5} (2 \sin t - \cos t) \right]_0^\pi$$ Evaluating at the limits: $$A = \frac{e^{2\pi}}{5} (2 \sin \pi - \cos \pi) - \frac{e^0}{5} (2 \sin 0 - \cos 0)$$ $$A = \frac{e^{2\pi}}{5} (0 - (-1)) - \frac{1}{5} (0 - 1)$$ $$A = \frac{1}{5} e^{2\pi} + \frac{1}{5}$$ ### Final Calculation Comparing our result $A = \frac{1}{5} e^{2\pi} + \frac{1}{5}$ with the given form $\alpha e^{2\pi} + \beta$, we identify: $$\alpha = \frac{1}{5}, \quad \beta = \frac{1}{5}$$ The final required value is: $$10(\alpha + \beta) = 10\left(\frac{1}{5} + \frac{1}{5}\right) = 10\left(\frac{2}{5}\right) = \mathbf{4}$$

JEE Main 2021 (18 March Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x d y - y d x = x^{2} - y^{2} d x, x \geq 1$ , with $y (1) = 0$ . If the area bounded by the line $x = 1, x = e^{π}, y = 0$ and $y = y (x)$ is $α e^{2 π} + β$ , then the value of $10 (α + β)$ is equal to ____.

Q: Can I use the substitution $y = vx$ instead of the quotient rule?

Absolutely! Substituting $y = vx$ is the standard method for homogeneous differential equations. If you set $y = vx$, then $dy = vdx + xdv$. Substituting this into the original equation will yield the same result. The quotient rule method is just a faster, more elegant shortcut that saves you precious time in the exam hall.

Q: Wait, why did the limits change from $1$ to $e^\pi$ to $0$ to $\pi$?

When we performed the substitution $t = \ln x$, we changed the variable of integration. Since the limits of integration are values of $x$, we must map them to the corresponding values of $t$. When $x=1$, $t = \ln(1) = 0$. When $x=e^\pi$, $t = \ln(e^\pi) = \pi$. Always update your limits immediately after substitution to avoid errors!

Q: Is it possible to get a negative area here?

In this specific problem, $y = x \sin(\ln x)$ is positive for $x \in [1, e^\pi]$ because $\ln x$ ranges from $0$ to $\pi$, and the sine function is non-negative in the interval $[0, \pi]$. Since $x$ is also positive, the area must be positive. If you ever calculate a negative area for a region bounded by a curve and the x-axis, double-check your integration limits or the sign of your function!

Enter Numerical Value:

Visualized Solution

The Differential Equation

Given: $x d y - y d x = x^{2} - y^{2} d x$
Initial condition: $y (1) = 0$

Manipulating the Equation

Divide both sides by $x^{2}$ :
$\frac{x d y - y d x}{x ^{2}} = \frac{x ^{2} - y ^{2}}{x ^{2}} d x$

Exact Differential Form

Recall quotient rule: $d (\frac{y}{x}) = \frac{x d y - y d x}{x ^{2}}$
Rewrite RHS: $\frac{x ^{2} - y ^{2}}{x ^{2}} = \frac{1 - ( y / x ) ^{2}}{x}$
Equation becomes: $d (\frac{y}{x}) = \frac{1 - ( y / x ) ^{2}}{x} d x$

Separating Variables

Rearrange to separate variables:
$\frac{d ( y / x )}{1 - ( y / x ) ^{2}} = \frac{d x}{x}$

Integration

Integrate: $\int \frac{d ( y / x )}{1 - ( y / x ) ^{2}} = \int \frac{d x}{x}$
Result: $sin^{- 1} (\frac{y}{x}) = ln x + C$

Applying Initial Conditions

Use $y (1) = 0$ :
$sin^{- 1} (\frac{0}{1}) = ln (1) + C$
$0 = 0 + C \Rightarrow C = 0$

The Curve Equation

$sin^{- 1} (\frac{y}{x}) = ln x$
$\frac{y}{x} = sin (ln x)$
$y = x sin (ln x)$

Visualizing the Area

Area $A = \int_{1}^{e^{π}} y (x) d x$
$A = \int_{1}^{e^{π}} x sin (ln x) d x$

Integral Substitution

Let $ln x = t \Rightarrow x = e^{t}$
Differentiating: $d x = e^{t} d t$
Limits: $x = 1 \Rightarrow t = 0$
Limits: $x = e^{π} \Rightarrow t = π$

Transformed Area Integral

Substitute into the area integral:
$A = \int_{0}^{π} e^{t} sin t \cdot e^{t} d t$
$A = \int_{0}^{π} e^{2 t} sin t d t$

Standard Integration Formula

Formula: $\int e^{a x} sin b x d x = \frac{e ^{a x}}{a ^{2} + b ^{2}} (a sin b x - b cos b x)$
Here $a = 2, b = 1$ :
$A = [\frac{e ^{2 t}}{2 ^{2} + 1 ^{2}} (2 sin t - cos t)]_{0}^{π}$

Evaluating Limits

Upper limit ( $t = π$ ): $\frac{e ^{2 π}}{5} (2 sin π - cos π) = \frac{e ^{2 π}}{5} (0 - (- 1)) = \frac{e ^{2 π}}{5}$
Lower limit ( $t = 0$ ): $\frac{e ^{0}}{5} (2 sin 0 - cos 0) = \frac{1}{5} (0 - 1) = - \frac{1}{5}$
$A = \frac{e ^{2 π}}{5} - (- \frac{1}{5}) = \frac{1}{5} e^{2 π} + \frac{1}{5}$

Comparing and Final Answer

Compare $A = \frac{1}{5} e^{2 π} + \frac{1}{5}$ with $α e^{2 π} + β$
$α = \frac{1}{5}, β = \frac{1}{5}$
Calculate $10 (α + β) = 10 (\frac{1}{5} + \frac{1}{5}) = 10 (\frac{2}{5}) = 4$

00:00 / 00:00

The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

We begin with the differential equation:

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

At first glance, this appears complex. However, we recognize that the term

x d y - y d x

is the numerator of the derivative of the quotient

\frac{y}{x}

By dividing both sides of the equation by

x^{2}

, we transform the expression into a recognizable form:

d (\frac{y}{x}) = \frac{1 - ( y / x ) ^{2}}{x} d x

The Elegance of Substitution

By separating the variables, we arrive at the following integral:

\int \frac{d ( y / x )}{1 - ( y / x ) ^{2}} = \int \frac{d x}{x}

The left side is the standard integral for

sin^{- 1} (\frac{y}{x})

, while the right side yields

ln x + C

. Applying the initial condition

y (1) = 0

, we find that

C = 0

This reveals the hidden curve:

y = x sin (ln x)

The Journey Through the Integral

We must find the area

A

bounded by this curve from

x = 1

x = e^{π}

A = \int_{1}^{e^{π}} x sin (ln x) d x

To solve this, we use the substitution

t = ln x

, which implies

x = e^{t}

and

d x = e^{t} d t

. The limits of integration change from

[1, e^{π}]

[0, π]

The integral becomes:

A = \int_{0}^{π} e^{2 t} sin t d t

Evaluating the Integral

We utilize the standard integration formula:

\int e^{a t} sin b t d t = \frac{e ^{a t}}{a ^{2} + b ^{2}} (a sin b t - b cos b t)

Applying this with

a = 2

and

b = 1

A = [\frac{e ^{2 t}}{5} (2 sin t - cos t)]_{0}^{π}

Evaluating at the limits:

A = \frac{e ^{2 π}}{5} (2 sin π - cos π) - \frac{e ^{0}}{5} (2 sin 0 - cos 0)

A = \frac{e ^{2 π}}{5} (0 - (- 1)) - \frac{1}{5} (0 - 1)

A = \frac{1}{5} e^{2 π} + \frac{1}{5}

Final Calculation

Comparing our result

A = \frac{1}{5} e^{2 π} + \frac{1}{5}

with the given form

α e^{2 π} + β

, we identify:

α = \frac{1}{5}, β = \frac{1}{5}

The final required value is:

10 (α + β) = 10 (\frac{1}{5} + \frac{1}{5}) = 10 (\frac{2}{5}) = 4

Common Traps & FAQs

Why did we divide by

x^{2}

at the start? How was I supposed to know that?

That is a brilliant question! In JEE problems, whenever you see terms like

x d y - y d x

, your intuition should immediately jump to the quotient rule for differentiation:

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

. By dividing by

x^{2}

, we force the equation into a form where we can integrate directly. It is a classic 'pattern recognition' move that turns a scary equation into a simple one.

Can I use the substitution

y = v x

instead of the quotient rule?

Absolutely! Substituting

y = v x

is the standard method for homogeneous differential equations. If you set

y = v x

, then

d y = v d x + x d v

. Substituting this into the original equation will yield the same result. The quotient rule method is just a faster, more elegant shortcut that saves you precious time in the exam hall.

Why does the integral

\int e^{2 t} sin t d t

result in a sum of terms?

When you integrate an exponential multiplied by a trigonometric function, the 'cyclic' nature of their derivatives means they never truly disappear. Using integration by parts twice (or the standard formula) creates a loop that eventually resolves into the expression

\frac{e ^{a t}}{a ^{2} + b ^{2}} (a sin b t - b cos b t)

. It is not a mistake; it is the beautiful, oscillating nature of the function!

Wait, why did the limits change from

1

e^{π}

0

π

When we performed the substitution

t = ln x

, we changed the variable of integration. Since the limits of integration are values of

x

, we must map them to the corresponding values of

t

. When

x = 1

t = ln (1) = 0

. When

x = e^{π}

t = ln (e^{π}) = π

. Always update your limits immediately after substitution to avoid errors!

Is it possible to get a negative area here?

In this specific problem,

y = x sin (ln x)

is positive for

x \in [1, e^{π}]

because

ln x

ranges from

0

π

, and the sine function is non-negative in the interval

[0, π]

. Since

x

is also positive, the area must be positive. If you ever calculate a negative area for a region bounded by a curve and the x-axis, double-check your integration limits or the sign of your function!

Enter Numerical Value:

Visualized Solution

The Differential Equation

Manipulating the Equation

Exact Differential Form

Separating Variables

Integration

Applying Initial Conditions

The Curve Equation

Visualizing the Area

Integral Substitution

Transformed Area Integral

Standard Integration Formula

Evaluating Limits

Comparing and Final Answer

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Elegance of Substitution

The Journey Through the Integral

Evaluating the Integral

Final Calculation

Similar Questions

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If $y \frac{d y}{d x} = x [\frac{y ^{2}}{x ^{2}} + \frac{ϕ ( \frac{y ^{2}}{x ^{2}} )}{ϕ ^{'} ( \frac{y ^{2}}{x ^{2}} )}], x > 0, ϕ > 0$ , and $y (1) = - 1$ , then $ϕ (\frac{y ^{2}}{4})$ is equal to :

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Enter Numerical Value:

Visualized Solution

The Differential Equation

Manipulating the Equation

Exact Differential Form

Separating Variables

Integration

Applying Initial Conditions

The Curve Equation

Visualizing the Area

Integral Substitution

Transformed Area Integral

Standard Integration Formula

Evaluating Limits

Comparing and Final Answer

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Elegance of Substitution

The Journey Through the Integral

Evaluating the Integral

Final Calculation

Similar Questions

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