Let y = f(x) is the solution of the differential equation ey (dy/dx - 1) = ex such that y(0) = 0, th

Q: Wait, why can't we just use separation of variables here? It looks like we have $e^y$ and $e^x$ terms.

That is a classic trap! If you try to separate the variables, you will find that the term $(\frac{dy}{dx} - 1)$ is coupled with $e^y$. You cannot isolate $dy$ and $dx$ without leaving a mixed term behind. The presence of the derivative inside the parenthesis makes this a non-separable equation, which is why we must use the substitution method to linearize it.

Q: Why is the integrating factor $e^{-x}$ and not $e^x$? I keep getting confused with the signs.

This is a common point of friction. When we transform the equation, we get $\frac{dt}{dx} - t = e^x$. Comparing this to the standard form $\frac{dt}{dx} + P(x)t = Q(x)$, we see that $P(x) = -1$. The formula for the Integrating Factor is $e^{\int P(x) dx}$. Therefore, we have $e^{\int -1 dx}$, which results in $e^{-x}$. Always be meticulous with the sign of $P(x)$!

Q: Why do we need to solve for the constant $C$ before finding $y(1)$?

The constant $C$ represents the family of all possible solution curves for this differential equation. Without the initial condition $y(0) = 0$, we would have an infinite number of curves. By using the initial condition, we 'pin' the solution to a specific curve that passes through the origin. Only then can we accurately predict the value of $y$ at $x = 1$.

Question

Let $y = f(x)$ is the solution of the differential equation $e^y \left(\frac{dy}{dx} - 1\right) = e^x$ such that $y(0) = 0$, then $y(1)$ is equal to:

Options:

(A) $\ln2$

(B) $2 + \ln2$

(C) $1 + \ln2$

(D) $3 + \ln2$

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we are going to dismantle a problem that often intimidates students in the JEE Advanced examination hall. It is a differential equation:

$$e^y \left( \frac{dy}{dx} - 1 \right) = e^x$$

Many students see this and immediately panic, trying to separate variables or force a solution that isn't there. In mathematics, when a problem looks overly complex, it is usually hiding a beautiful, simple structure underneath.

### The Art of Substitution

Look at the term $e^y$ and the term $\frac{dy}{dx}$. If we differentiate $e^y$ with respect to $x$, the chain rule gives us $e^y \frac{dy}{dx}$.

If we let $t = e^y$, then:

$$\frac{dt}{dx} = e^y \frac{dy}{dx}$$

This substitution is our masterstroke. It transforms a non-linear, intimidating equation into a standard, friendly Linear Differential Equation.

### The Linear Transformation

Let us expand the original equation:

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

Now, substitute our new variable $t$. The first term becomes $\frac{dt}{dx}$, and the second term becomes $t$. The equation simplifies to:

$$\frac{dt}{dx} - t = e^x$$

We have moved from a complex, non-linear world into the structured, predictable world of Linear Differential Equations. This is the standard form $\frac{dt}{dx} + P(x)t = Q(x)$, where $P(x) = -1$ and $Q(x) = e^x$.

### The Integrating Factor

To solve any linear differential equation, we need the engine that drives the solution: the Integrating Factor (I.F.). The formula is:

$$I.F. = e^{\int P(x) dx}$$

Here, $P(x) = -1$, so our Integrating Factor is:

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

When we multiply our entire equation $\frac{dt}{dx} - t = e^x$ by $e^{-x}$, the left side becomes the derivative of the product of our dependent variable and the Integrating Factor:

$$\frac{d}{dx} (t \cdot e^{-x}) = e^x \cdot e^{-x} = e^0 = 1$$

### The Final Integration and Boundary Condition

Now, we integrate both sides with respect to $x$. The integral of $1$ is simply $x$, plus our constant of integration $C$:

$$t \cdot e^{-x} = x + C$$

Substituting back $t = e^y$, we get:

$$e^y \cdot e^{-x} = x + C \Rightarrow e^{y-x} = x + C$$

Using the initial condition $y(0) = 0$, we substitute $x = 0$ and $y = 0$:

$$e^{0-0} = 0 + C \Rightarrow 1 = C$$

Our specific solution is $e^{y-x} = x + 1$. To find $y(1)$, we take the natural logarithm of both sides:

$$y - x = \ln(x + 1) \Rightarrow y = x + \ln(x + 1)$$

Finally, substituting $x = 1$:

$$y(1) = 1 + \ln(1 + 1) = **1 + \ln 2**$$

The destination of our journey is exactly **$1 + \ln 2$**. Keep this mindset for your JEE preparation: look for the substitution, identify the standard form, and trust the process.

Animated Solution for Mathematics - Differential Equations: Let $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

Select Answer:

Visualized Solution

Analyze the Equation Structure

Substitution Strategy $e^{y} = t$

Transforming the Equation

Identifying $P (x)$ and $Q (x)$

Calculating Integrating Factor

Setting up the General Solution

Integrating the Right-Hand Side

Back-Substitution of $t = e^{y}$

Applying Initial Condition $y (0) = 0$

Finding the Explicit Function $y (x)$

Calculating $y (1)$

The Sigma Insight: Linear Differential Equations

The Art of Substitution

The Linear Transformation

The Integrating Factor

The Final Integration and Boundary Condition

Similar Questions

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

Select Answer:

Visualized Solution

Analyze the Equation Structure

Substitution Strategy .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ey=t

Transforming the Equation

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(x)

Calculating Integrating Factor

Setting up the General Solution

Integrating the Right-Hand Side

Back-Substitution of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=ey

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

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

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

The Sigma Insight: Linear Differential Equations

The Art of Substitution

The Linear Transformation

The Integrating Factor

The Final Integration and Boundary Condition

Similar Questions

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