Suppose y = y(x) be the solution curve to the differential equation dy/dx - y = 2 - e⁻ˣ such that li

Q: Wait, why do we reject the $Ce^x$ term? Can't it just be a constant?

That is a fantastic question! The condition $\lim_{x \to \infty} y(x)$ is finite is the constraint that forces our hand. As $x \to \infty$, the term $e^x$ grows without bound. If $C$ were any non-zero value, $y(x)$ would approach either positive or negative infinity. Therefore, the only way to satisfy the condition is to set $C=0$, effectively 'killing' the term that causes the divergence.

Q: Why is the integrating factor $e^{-x}$? I thought it was $e^x$.

Remember the standard form of the linear differential equation: $\frac{dy}{dx} + Py = Q$. In our equation, $\frac{dy}{dx} - y = 2 - e^{-x}$, the coefficient $P$ is $-1$. The integrating factor is defined as $I.F. = e^{\int P dx}$. Since $\int -1 dx = -x$, the integrating factor is indeed $e^{-x}$. Always be careful with the sign of $P$!

Q: Is there a trick to finding the intercepts $a$ and $b$?

No tricks, just the fundamental definitions! The $x$-intercept $a$ is simply the value of $x$ when $y=0$. The $y$-intercept $b$ is the value of $y$ when $x=0$. Once you have the linear equation of the tangent, just plug in $0$ for the other variable and solve. It is a straightforward algebraic finish to a beautiful calculus problem.

Question

Suppose $y = y(x)$ be the solution curve to the differential equation $\frac{dy}{dx} - y = 2 - e^{-x}$ such that $\lim_{x 	o \infty} y(x)$ is finite. If $a$ and $b$ are respectively the x- and y- intercepts of the tangent to the curve at $x=0$, then the value of $a-4b$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of the Differential Equation

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a differential equation; we are uncovering the hidden geometry of a curve.

We are given the equation:

$$ \frac{dy}{dx} - y = 2 - e^{-x} $$

We are also provided with a mysterious constraint: the curve must remain finite as $x$ marches toward infinity. Let us embark on this journey step by step.

### Phase 1: The Anatomy of the Equation

First, we must recognize the beast we are facing. This is a classic **Linear Differential Equation** of the form:

$$ \frac{dy}{dx} + Py = Q $$

By comparing our equation $\frac{dy}{dx} - y = 2 - e^{-x}$ to the standard form, we immediately identify that $P = -1$ and $Q = 2 - e^{-x}$.

This identification is the foundation of our entire solution. Without it, we are lost; with it, we have a roadmap.

### Phase 2: The Magic of the Integrating Factor

To solve this, we need a tool that transforms the left side into a perfect derivative. This tool is the Integrating Factor (I.F.).

We calculate it as:

$$ I.F. = e^{\int P \, dx} $$

Substituting our $P = -1$, we get:

$$ I.F. = e^{\int -1 \, dx} = e^{-x} $$

This $e^{-x}$ is the key that unlocks the equation. When we multiply the entire differential equation by this factor, the left side magically becomes the derivative of the product $y \cdot e^{-x}$. It is one of the most elegant moments in calculus.

### Phase 3: The General Solution and the Limit Trap

Now, we integrate both sides:

$$ y \cdot e^{-x} = \int (2 - e^{-x}) e^{-x} \, dx + C $$

Expanding the integral, we get:

$$ \int (2e^{-x} - e^{-2x}) \, dx $$

Integrating term by term, we arrive at:

$$ y \cdot e^{-x} = -2e^{-x} + \frac{1}{2}e^{-2x} + C $$

Multiplying by $e^x$ to isolate $y$, we find the general solution:

$$ y = -2 + \frac{1}{2}e^{-x} + Ce^x $$

Here is where the JEE examiners test your conceptual depth. We are told $\lim_{x 	o \infty} y(x)$ is finite.

Look at our solution. As $x 	o \infty$, $e^{-x}$ vanishes to zero, but $e^x$ explodes to infinity. If $C$ is anything other than zero, our curve will blow up.

Thus, the condition forces $C=0$. We are left with the specific, elegant curve:

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

### Phase 4: Geometry Meets Calculus

We have our curve. Now, we need the tangent at $x=0$.

First, the point of tangency:

$$ y(0) = -2 + \frac{1}{2}e^0 = -2 + 0.5 = -1.5 $$

So, our point is $(0, -1.5)$. Next, we find the slope. We differentiate $y(x) = -2 + \frac{1}{2}e^{-x}$ to get:

$$ \frac{dy}{dx} = -\frac{1}{2}e^{-x} $$

At $x=0$, the slope $m = -\frac{1}{2}$.

Using the point-slope form $y - y_1 = m(x - x_1)$, we write:

$$ y - (-1.5) = -0.5(x - 0) $$

This simplifies to $y + 1.5 = -0.5x$. Multiplying by $2$, we get $2y + 3 = -x$, or:

$$ x + 2y + 3 = 0 $$

This is the equation of our tangent line.

### Phase 5: The Final Act

Finally, we find the intercepts. For the $x$-intercept $a$, set $y=0$:

$$ x + 2(0) + 3 = 0 \implies a = -3 $$

For the $y$-intercept $b$, set $x=0$:

$$ 0 + 2y + 3 = 0 \implies y = -1.5, 	ext{ so } b = -1.5 $$

The problem asks for $a - 4b$. Substituting our values:

$$ -3 - 4(-1.5) = -3 + 6 = 3 $$

And there it is. The complexity collapses into the simple, beautiful integer **3**. You have navigated the differential equation, respected the limit condition, applied the geometry of tangents, and arrived at the truth. Well done!

Enter Numerical Value:

Visualized Solution

Identifying the Linear Differential Equation

Calculating the Integrating Factor (I.F.)

Setting up the General Solution

Solving the Integral

Finding the Explicit Form of $y (x)$

Applying the Limit Condition

Finding the Point of Tangency at $x = 0$

Finding the Slope of the Tangent

Equation of the Tangent Line

Finding Intercepts $a$ and $b$

Final Calculation: $a - 4 b$

The Sigma Insight: Linear Differential Equations

The Dance of the Differential Equation

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

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Enter Numerical Value:

Visualized Solution

Identifying the Linear Differential Equation

Calculating the Integrating Factor (I.F.)

Setting up the General Solution

Solving the Integral

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

Applying the Limit Condition

Finding the Point of Tangency at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Finding the Slope of the Tangent

Equation of the Tangent Line

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a−4b