Let $y = y(x)$ be the solution of the differential equation $\left((x + 2)e^{\frac{y+1}{x+2}} + (y + 1)\right) dx = (x + 2)dy, y(1) = 1$. If the domain of $y = y(x)$ is an open interval $(\alpha, \beta)$, then $|\alpha + \beta|$ is equal to ____.

### Analyzing the Setup The given differential equation is: $$((x + 2)e^{\frac{y+1}{x+2}} + (y + 1)) dx = (x + 2)dy$$ The repetition of the terms $(x+2)$ and $(y+1)$ suggests a change of variables. Let us define: $X = x+2$ and $Y = y+1$ Differentiating these, we obtain $dX = dx$ and $dY = dy$. Substituting these into the original equation, we get: $$(X e^{\frac{Y}{X}} + Y) dX = X dY$$ ### The Hidden Quotient Rule Rearranging the terms to isolate the differential components, we have: $$X e^{\frac{Y}{X}} dX = X dY - Y dX$$ Observe that the right side, $X dY - Y dX$, is the numerator of the derivative of a quotient. Dividing the entire equation by $X^2$ yields: $$\frac{e^{\frac{Y}{X}}}{X} dX = \frac{X dY - Y dX}{X^2}$$ The right side is now the exact differential of the quotient $\frac{Y}{X}$. We can rewrite the equation as: $$\frac{1}{X} dX = e^{-\frac{Y}{X}} d\left(\frac{Y}{X}\right)$$ ### Integration and the Constant Integrating both sides of the equation: $$\int \frac{1}{X} dX = \int e^{-\frac{Y}{X}} d\left(\frac{Y}{X}\right)$$ This results in the following relation: $$\ln|X| = -e^{-\frac{Y}{X}} + C$$ To determine the constant $C$, we use the initial condition $y(1)=1$. In our transformed coordinates, this corresponds to $X = 1+2 = 3$ and $Y = 1+1 = 2$. Substituting these values: $$\ln 3 = -e^{-\frac{2}{3}} + C \implies C = \ln 3 + e^{-\frac{2}{3}}$$ ### Final Calculation and Domain Analysis Substituting $C$ back into our general solution, we obtain: $$\ln|X| = -e^{-\frac{Y}{X}} + \ln 3 + e^{-\frac{2}{3}}$$ Rearranging for the exponential term gives: $$e^{-\frac{Y}{X}} = \ln 3 + e^{-\frac{2}{3}} - \ln|X|$$ For $y$ to be defined, the right side must be positive, leading to the inequality: $$\ln|X| < \ln 3 + e^{-\frac{2}{3}}$$ Taking the exponential of both sides, we find $|X| < 3e^{e^{-\frac{2}{3}}}$. Let $k = 3e^{e^{-\frac{2}{3}}}$. Since $|x+2| < k$, the domain is defined by: $$-k-2 < x < k-2$$ Thus, the interval is $(\alpha, \beta) = (-k-2, k-2)$. The sum of the endpoints is: $$\alpha + \beta = (-k-2) + (k-2) = -4$$ The final result is: $$|\alpha + \beta| = \mathbf{4}$$

JEE Main 2021 (22 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $((x + 2) e^{\frac{y + 1}{x + 2}} + (y + 1)) d x = (x + 2) d y, y (1) = 1$ . If the domain of $y = y (x)$ is an open interval $(α, β)$ , then $∣ α + β ∣$ is equal to ____.

Q: Why did we substitute $X = x+2$ and $Y = y+1$? Is there a faster way?

Great question! While you could try other methods, the recurring linear blocks $(x+2)$ and $(y+1)$ are a classic signal in JEE problems. By shifting the origin to $(-2, -1)$, we transform a complex-looking equation into a homogeneous one, which is much easier to handle. It is not just about speed; it is about reducing the cognitive load and avoiding algebraic errors.

Q: How did you know to divide by $X^2$ in the second step?

That is the 'Aha!' moment of the problem. We noticed the expression $X dY - Y dX$ on the right side. This is the numerator of the quotient rule for the derivative of $\frac{Y}{X}$. Dividing by $X^2$ completes the differential $d(\frac{Y}{X}) = \frac{X dY - Y dX}{X^2}$. Recognizing these patterns is a key skill in mastering differential equations.

Q: Why is the domain restricted by the RHS being positive?

The equation we derived is $e^{-\frac{Y}{X}} = \text{RHS}$. Since the exponential function $e^u$ is always strictly positive for any real $u$, the expression on the right-hand side must also be strictly positive for the equation to hold. This constraint is what defines the boundaries of our domain $(\alpha, \beta)$.

Enter Numerical Value:

Visualized Solution

Identifying the Structure of the DE

Given DE: $((x + 2) e^{\frac{y + 1}{x + 2}} + (y + 1)) d x = (x + 2) d y$
Notice the recurring terms $(x + 2)$ and $(y + 1)$ .
Let's substitute: $X = x + 2$ and $Y = y + 1$ .
Differentiating both gives: $d X = d x$ and $d Y = d y$ .

Transforming to $X$ and $Y$

Substitute $X$ and $Y$ into the DE:
$(X e^{\frac{Y}{X}} + Y) d X = X d Y$

The Quotient Rule Trick

Rearrange the terms: $X e^{\frac{Y}{X}} d X = X d Y - Y d X$
Divide both sides by $X^{2}$ to recognize the quotient rule derivative:
$\frac{e ^{\frac{Y}{X}}}{X} d X = \frac{X d Y - Y d X}{X ^{2}}$

Recognizing $d (\frac{Y}{X})$

Recall that $d (\frac{Y}{X}) = \frac{X d Y - Y d X}{X ^{2}}$ .
The equation becomes: $\frac{1}{X} d X = e^{- \frac{Y}{X}} d (\frac{Y}{X})$

Integration of Both Sides

Integrate both sides: $\int \frac{1}{X} d X = \int e^{- \frac{Y}{X}} d (\frac{Y}{X})$
Result: $ln ∣ X ∣ = - e^{- \frac{Y}{X}} + C$

Applying $y (1) = 1$

Initial condition: $y (1) = 1$ .
In terms of $X$ and $Y$ : $x = 1 \Rightarrow X = 3$ , $y = 1 \Rightarrow Y = 2$ .
Substitute into the solution: $ln 3 = - e^{- \frac{2}{3}} + C$
Find $C$ : $C = ln 3 + e^{- \frac{2}{3}}$

The General Solution

Substitute $C$ back: $ln ∣ x + 2∣ = - e^{- \frac{y + 1}{x + 2}} + ln 3 + e^{- \frac{2}{3}}$
Rearrange for the exponential term:
$e^{- \frac{y + 1}{x + 2}} = ln 3 + e^{- \frac{2}{3}} - ln ∣ x + 2∣$

Finding the Domain of $y (x)$

For $y$ to be defined as a real function, the RHS must be positive:
$ln 3 + e^{- \frac{2}{3}} - ln ∣ x + 2∣ > 0$
$ln ∣ x + 2∣ < ln 3 + e^{- \frac{2}{3}}$

Solving the Inequality for $x$

Take the exponential of both sides:
$∣ x + 2∣ < e^{l n 3 + e^{- \frac{2}{3}}} = 3 e^{e^{- \frac{2}{3}}}$
Let $k = 3 e^{e^{- \frac{2}{3}}}$ . Then $∣ x + 2∣ < k$ .
This implies $- k < x + 2 < k \Rightarrow - k - 2 < x < k - 2$ .

Identifying $α$ and $β$

The domain is $(α, β) = (- k - 2, k - 2)$ .
Calculate $α + β$ :
$α + β = (- k - 2) + (k - 2) = - 4$

Final Conclusion: $∣ α + β ∣$

Final step: $∣ α + β ∣ = ∣ - 4∣ = 4$ .
Key Takeaway: Shifting the origin using linear substitutions can simplify complex differential equations.
Next Challenge: Try solving the same DE with a different initial condition, like $y (0) = 0$ , and see how the domain changes.

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The given differential equation is:

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

The repetition of the terms

(x + 2)

and

(y + 1)

suggests a change of variables. Let us define:

X = x + 2

and

Y = y + 1

Differentiating these, we obtain

d X = d x

and

d Y = d y

. Substituting these into the original equation, we get:

(X e^{\frac{Y}{X}} + Y) d X = X d Y

The Hidden Quotient Rule

Rearranging the terms to isolate the differential components, we have:

X e^{\frac{Y}{X}} d X = X d Y - Y d X

Observe that the right side,

X d Y - Y d X

, is the numerator of the derivative of a quotient. Dividing the entire equation by

X^{2}

yields:

\frac{e ^{\frac{Y}{X}}}{X} d X = \frac{X d Y - Y d X}{X ^{2}}

The right side is now the exact differential of the quotient

\frac{Y}{X}

. We can rewrite the equation as:

\frac{1}{X} d X = e^{- \frac{Y}{X}} d (\frac{Y}{X})

Integration and the Constant

Integrating both sides of the equation:

\int \frac{1}{X} d X = \int e^{- \frac{Y}{X}} d (\frac{Y}{X})

This results in the following relation:

ln ∣ X ∣ = - e^{- \frac{Y}{X}} + C

To determine the constant

C

, we use the initial condition

y (1) = 1

. In our transformed coordinates, this corresponds to

X = 1 + 2 = 3

and

Y = 1 + 1 = 2

. Substituting these values:

ln 3 = - e^{- \frac{2}{3}} + C ⟹ C = ln 3 + e^{- \frac{2}{3}}

Final Calculation and Domain Analysis

Substituting

C

back into our general solution, we obtain:

ln ∣ X ∣ = - e^{- \frac{Y}{X}} + ln 3 + e^{- \frac{2}{3}}

Rearranging for the exponential term gives:

e^{- \frac{Y}{X}} = ln 3 + e^{- \frac{2}{3}} - ln ∣ X ∣

For

y

to be defined, the right side must be positive, leading to the inequality:

ln ∣ X ∣ < ln 3 + e^{- \frac{2}{3}}

Taking the exponential of both sides, we find

∣ X ∣ < 3 e^{e^{- \frac{2}{3}}}

. Let

k = 3 e^{e^{- \frac{2}{3}}}

. Since

∣ x + 2∣ < k

, the domain is defined by:

- k - 2 < x < k - 2

Thus, the interval is

(α, β) = (- k - 2, k - 2)

. The sum of the endpoints is:

α + β = (- k - 2) + (k - 2) = - 4

The final result is:

∣ α + β ∣ = 4

Common Traps & FAQs

Why did we substitute

X = x + 2

and

Y = y + 1

? Is there a faster way?

Great question! While you could try other methods, the recurring linear blocks

(x + 2)

and

(y + 1)

are a classic signal in JEE problems. By shifting the origin to

(- 2, - 1)

, we transform a complex-looking equation into a homogeneous one, which is much easier to handle. It is not just about speed; it is about reducing the cognitive load and avoiding algebraic errors.

How did you know to divide by

X^{2}

in the second step?

That is the 'Aha!' moment of the problem. We noticed the expression

X d Y - Y d X

on the right side. This is the numerator of the quotient rule for the derivative of

\frac{Y}{X}

. Dividing by

X^{2}

completes the differential

d (\frac{Y}{X}) = \frac{X d Y - Y d X}{X ^{2}}

. Recognizing these patterns is a key skill in mastering differential equations.

Why is the domain restricted by the RHS being positive?

The equation we derived is

e^{- \frac{Y}{X}} = RHS

. Since the exponential function

e^{u}

is always strictly positive for any real

u

, the expression on the right-hand side must also be strictly positive for the equation to hold. This constraint is what defines the boundaries of our domain

(α, β)

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $((x + 2) e^{\frac{y + 1}{x + 2}} + (y + 1)) d x = (x + 2) d y, y (1) = 1$ . If the domain of $y = y (x)$ is an open interval $(α, β)$ , then $∣ α + β ∣$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Identifying the Structure of the DE

Transforming to $X$ and $Y$

The Quotient Rule Trick

Recognizing $d (\frac{Y}{X})$

Integration of Both Sides

Applying $y (1) = 1$

The General Solution

Finding the Domain of $y (x)$

Solving the Inequality for $x$

Identifying $α$ and $β$

Final Conclusion: $∣ α + β ∣$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Hidden Quotient Rule

Integration and the Constant

Final Calculation and Domain Analysis

Similar Questions

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

Let the curve $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} = 2 (x + 1)$ . If the numerical value of area bounded by the curve $y = y (x)$ and $x$ -axis is $\frac{4 8}{3}$ , then the value of $y (1)$ is equal to ____

If the solution of the differential equation $(2 x + 3 y - 2) d x + (4 x + 6 y - 7) d y = 0, y (0) = 3$ , is $α x + β y + 3 lo g_{e} ∣2 x + 3 y - γ ∣ = 6$ , then $α + 2 β + 3 γ$ is equal to_____

If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ is equal to

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 ______.

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

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$

Enter Numerical Value:

Visualized Solution

Identifying the Structure of the DE

Transforming to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y

The Quotient Rule Trick

Recognizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d(XY​)

Integration of Both Sides

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(1)=1

The General Solution

Finding the Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

Solving the Inequality for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣α+β∣

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Hidden Quotient Rule

Integration and the Constant

Final Calculation and Domain Analysis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(3/2)α−tan(3/2)​, then α+2β is

.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 dy=eαx+ydx;α∈N. If y(loge​2)=loge​2 and y(0)=loge​(1/2), then the value of α 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 solution of the differential equation loge​(dxdy​)=3x+4y, with y(0)=0. If y(−32​loge​2)=αloge​2, then the value of α is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution of the differential equation (2x+3y−2)dx+(4x+6y−7)dy=0,y(0)=3, is αx+βy+3loge​∣2x+3y−γ∣=6, then α+2β+3γ 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),y∈[0,π/2) is the solution of the differential equation secydxdy​−sin(x+y)−sin(x−y)=0, with y(0)=0, then 5y′(2π​) 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; } Which of the following is true for y(x) that satisfies the differential equation dxdy​=xy−1+x−y;y(0)=0

Transforming to $X$ and $Y$

Recognizing $d (\frac{Y}{X})$

Applying $y (1) = 1$

Finding the Domain of $y (x)$

Solving the Inequality for $x$

Identifying $α$ and $β$

Final Conclusion: $∣ α + β ∣$

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

If the solution of the differential equation $(2 x + 3 y - 2) d x + (4 x + 6 y - 7) d y = 0, y (0) = 3$ , is $α x + β y + 3 lo g_{e} ∣2 x + 3 y - γ ∣ = 6$ , then $α + 2 β + 3 γ$ is equal to_____

If $y = y (x), y \in [0, π /2)$ is the solution of the differential equation $sec y \frac{d y}{d x} - sin (x + y) - sin (x - y) = 0$ , with $y (0) = 0$ , then $5 y^{'} (\frac{π}{2})$ 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

Which of the following is true for $y (x)$ that satisfies the differential equation $\frac{d y}{d x} = x y - 1 + x - y; y (0) = 0$