JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: If the solution curve, of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passing through the point $(2, 1)$ is $tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣$ , then $5 β + α$ is equal to

Q: Wait, why do we shift the origin to $(1, 1)$? Can't we just use $y=vx$ directly?

That is a great question! If you try to substitute $y=vx$ directly into the original equation $\frac{dy}{dx} = \frac{x+y-2}{x-y}$, you will find that the constant term $-2$ prevents the $x$ terms from canceling out completely. The equation is non-homogeneous because of that constant. By shifting the origin to the intersection point $(1, 1)$, we effectively 'center' the coordinate system, making the equation homogeneous and allowing the standard $Y=vX$ substitution to work perfectly.

Q: Why does the constant of integration $C$ turn out to be zero?

The constant $C$ is determined by the boundary condition provided: the curve passes through $(2, 1)$. When we substitute $x=2$ and $y=1$ into our derived equation, we get $X=1$ and $Y=0$, which implies $v=0$. Plugging these into $\tan^{-1}(v) - \frac{1}{2} \ln(1+v^2) = \ln|X| + C$, we get $\tan^{-1}(0) - \frac{1}{2} \ln(1) = \ln(1) + C$, which simplifies to $0 - 0 = 0 + C$. Thus, $C=0$ is a direct consequence of the specific point the curve passes through.

Q: How do I know for sure that $\alpha=1$ and $\beta=2$?

After integrating and substituting back $v = \frac{y-1}{x-1}$ and $X = x-1$, you get $\tan^{-1}(\frac{y-1}{x-1}) - \frac{1}{2} \ln(1 + (\frac{y-1}{x-1})^2) = \ln|x-1|$. By comparing this result term-by-term with the given form $\tan^{-1}(\frac{y-1}{x-1}) - \frac{1}{\beta} \ln(\alpha + (\frac{y-1}{x-1})^2) = \ln|x-1|$, you can see that $\frac{1}{\beta} = \frac{1}{2}$ (so $\beta=2$) and the constant inside the logarithm $\alpha$ must be $1$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Observe the structure of the differential equation

\frac{d y}{d x} = \frac{x + y - 2}{x - y}

. Notice that the numerator and denominator are linear expressions. If you shift the origin to the point of intersection of the lines

x + y - 2 = 0

and

x - y = 0

, the constant terms will vanish, transforming this into a homogeneous equation.

Hint 2

Hint 2 (The Arsenal): Once you have shifted the origin to

(1, 1)

using the substitution

X = x - 1

and

Y = y - 1

, the equation becomes

\frac{d Y}{d X} = \frac{X + Y}{X - Y}

. Recall that for a homogeneous equation of the form

\frac{d Y}{d X} = f (\frac{Y}{X})

, the standard substitution is

Y = v X

, which leads to a separable differential equation in terms of

v

and

X

Hint 3

Hint 3 (The Execution): After substituting

Y = v X

and simplifying, you will arrive at the separable form

\frac{1 - v}{1 + v ^{2}} d v = \frac{d X}{X}

. Integrate both sides, remembering to split the left-hand side integral into

\int \frac{1}{1 + v ^{2}} d v - \int \frac{v}{1 + v ^{2}} d v

. Finally, use the boundary condition

(2, 1)

to determine the constant of integration

C

Visualized Solution

Analyze the Differential Equation

Given DE: $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$
This is a non-homogeneous differential equation of the form $\frac{d y}{d x} = \frac{a x + b y + c}{d x + ey + f}$ .

Shift of Origin to $(1, 1)$

Solve $x + y - 2 = 0$ and $x - y = 0$ to find the intersection point.
Intersection: $(1, 1)$ .
Let $x - 1 = X$ and $y - 1 = Y$ .
Then $d x = d X$ and $d y = d Y$ , so $\frac{d y}{d x} = \frac{d Y}{d X}$ .

Transform to Homogeneous Form

Substitute $x = X + 1$ and $y = Y + 1$ into the DE:
$\frac{d Y}{d X} = \frac{( X + 1 ) + ( Y + 1 ) - 2}{( X + 1 ) - ( Y + 1 )}$
$\frac{d Y}{d X} = \frac{X + Y}{X - Y}$

Homogeneous Substitution $Y = v X$

Let $Y = v X$ .
Differentiating with respect to $X$ : $\frac{d Y}{d X} = v + X \frac{d v}{d X}$ .
Substitute into the DE: $v + X \frac{d v}{d X} = \frac{X + v X}{X - v X}$

Simplify the Expression

$v + X \frac{d v}{d X} = \frac{1 + v}{1 - v}$
$X \frac{d v}{d X} = \frac{1 + v}{1 - v} - v$
$X \frac{d v}{d X} = \frac{1 + v - v ( 1 - v )}{1 - v} = \frac{1 + v ^{2}}{1 - v}$

Separate Variables and Integrate

Rearrange to separate variables: $\frac{1 - v}{1 + v ^{2}} d v = \frac{d X}{X}$
Integrate both sides: $\int \frac{1 - v}{1 + v ^{2}} d v = \int \frac{d X}{X}$

Execute Integration

Split the LHS integral: $\int \frac{1}{1 + v ^{2}} d v - \int \frac{v}{1 + v ^{2}} d v = \int \frac{1}{X} d X$
$tan^{- 1} v - \frac{1}{2} ln (1 + v^{2}) = ln ∣ X ∣ + C$

Apply Boundary Condition $(2, 1)$

At point $(x, y) = (2, 1)$ :
$X = 2 - 1 = 1$ and $Y = 1 - 1 = 0$
$v = \frac{Y}{X} = \frac{0}{1} = 0$
Substitute into the solution: $tan^{- 1} (0) - \frac{1}{2} ln (1 + 0^{2}) = ln ∣1∣ + C$
$0 - 0 = 0 + C \Rightarrow C = 0$

Re-substitute and Compare

Particular solution: $tan^{- 1} v - \frac{1}{2} ln (1 + v^{2}) = ln ∣ X ∣$
Substitute $v = \frac{y - 1}{x - 1}$ and $X = x - 1$ :
$tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} ln (1 + (\frac{y - 1}{x - 1})^{2}) = ln ∣ x - 1∣$
Comparing with $tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} lo g_{e} (α + (\frac{y - 1}{x - 1})^{2}) = lo g_{e} ∣ x - 1∣$ :
$α = 1, β = 2$

Final Calculation

Calculate $5 β + α$ :
$5 (2) + 1 = 10 + 1 = 11$
Final Answer: $11$

00:00 / 00:00

The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

\frac{d y}{d x} = \frac{x + y - 2}{x - y}

This is a non-homogeneous differential equation because of the constant term

- 2

. To solve this, we must shift the origin to the point of intersection of the lines

x + y - 2 = 0

and

x - y = 0

Solving these simultaneous equations yields the point

(1, 1)

. We define new coordinates

X = x - 1

and

Y = y - 1

. This transformation simplifies the equation into the homogeneous form:

\frac{d Y}{d X} = \frac{X + Y}{X - Y}

The Elegant Substitution

To solve the homogeneous equation, we use the standard substitution

Y = v X

. Differentiating with respect to

X

using the product rule gives:

\frac{d Y}{d X} = v + X \frac{d v}{d X}

Substituting this into our equation, we obtain:

v + X \frac{d v}{d X} = \frac{X + v X}{X - v X} = \frac{1 + v}{1 - v}

Subtracting

v

from both sides and simplifying, we get:

X \frac{d v}{d X} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v ( 1 - v )}{1 - v} = \frac{1 + v ^{2}}{1 - v}

The Integration Journey

We have now arrived at a separable differential equation:

\frac{1 - v}{1 + v ^{2}} d v = \frac{d X}{X}

Integrating both sides, we split the left integral:

\int \frac{1}{1 + v ^{2}} d v - \int \frac{v}{1 + v ^{2}} d v = \int \frac{1}{X} d X

This yields the following result:

tan^{- 1} (v) - \frac{1}{2} ln (1 + v^{2}) = ln ∣ X ∣ + C

The Final Reveal

Given that the curve passes through

(2, 1)

, we have

X = 2 - 1 = 1

and

Y = 1 - 1 = 0

, which implies

v = 0

. Substituting these values into our integrated equation, we find

C = 0

Substituting

v = \frac{y - 1}{x - 1}

and

X = x - 1

back into the equation, we obtain:

tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} ln (1 + (\frac{y - 1}{x - 1})^{2}) = ln ∣ x - 1∣

By comparing this result to the required form, we identify the constants

α = 1

and

β = 2

. The final calculation is:

$5 β + α = 5 (2) + 1 = 11$

Common Traps & FAQs

Wait, why do we shift the origin to

(1, 1)

? Can't we just use

y = v x

directly?

That is a great question! If you try to substitute

y = v x

directly into the original equation

\frac{d y}{d x} = \frac{x + y - 2}{x - y}

, you will find that the constant term

- 2

prevents the

x

terms from canceling out completely. The equation is non-homogeneous because of that constant. By shifting the origin to the intersection point

(1, 1)

, we effectively 'center' the coordinate system, making the equation homogeneous and allowing the standard

Y = v X

substitution to work perfectly.

Why does the constant of integration

C

turn out to be zero?

The constant

C

is determined by the boundary condition provided: the curve passes through

(2, 1)

. When we substitute

x = 2

and

y = 1

into our derived equation, we get

X = 1

and

Y = 0

, which implies

v = 0

. Plugging these into

tan^{- 1} (v) - \frac{1}{2} ln (1 + v^{2}) = ln ∣ X ∣ + C

, we get

tan^{- 1} (0) - \frac{1}{2} ln (1) = ln (1) + C

, which simplifies to

0 - 0 = 0 + C

. Thus,

C = 0

is a direct consequence of the specific point the curve passes through.

How do I know for sure that

α = 1

and

β = 2

After integrating and substituting back

v = \frac{y - 1}{x - 1}

and

X = x - 1

, you get

tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{2} ln (1 + (\frac{y - 1}{x - 1})^{2}) = ln ∣ x - 1∣

. By comparing this result term-by-term with the given form

tan^{- 1} (\frac{y - 1}{x - 1}) - \frac{1}{β} ln (α + (\frac{y - 1}{x - 1})^{2}) = ln ∣ x - 1∣

, you can see that

\frac{1}{β} = \frac{1}{2}

(so

β = 2

) and the constant inside the logarithm

α

must be

1

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Shift of Origin to $(1, 1)$

Transform to Homogeneous Form

Homogeneous Substitution $Y = v X$

Simplify the Expression

Separate Variables and Integrate

Execute Integration

Apply Boundary Condition $(2, 1)$

Re-substitute and Compare

Final Calculation

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Elegant Substitution

The Integration Journey

The Final Reveal

Similar Questions

Let the solution curve $y = y (x)$ of the differential equation, $[\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{x ^{2} - y ^{2}} + e^{y / x}] y$ pass through the points $(1, 0)$ and $(2 α, α), α > 0$ . Then $α$ is equal to

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

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If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

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 ____.

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

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Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Shift of Origin to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1)

Transform to Homogeneous Form

Homogeneous Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y=vX

Simplify the Expression

Separate Variables and Integrate

Execute Integration

Apply Boundary Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2,1)

Re-substitute and Compare

Final Calculation

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Elegant Substitution

The Integration Journey

The Final Reveal

Similar Questions

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Shift of Origin to $(1, 1)$

Homogeneous Substitution $Y = v X$

Apply Boundary Condition $(2, 1)$

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

If $sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}$ is the solution of the differential equation $x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x$ and $y (1) = \frac{π}{3}$ , then $α^{2}$ is equal to

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

Let the solution curve of the differential equation $x \frac{d y}{d x} - y = y^{2} + 16 x^{2}, y (1) = 3$ be $y = y (x)$ . Then $y (2)$ is equal to :