If dy/dx = (2^(x+y) - 2^x)/2^y, y(0) = 1, then y(1) is equal to : (1) log2(2 + e), (2) log2(1 + e),

Q: Why can't I just integrate $2^y$ and $2^y - 1$ separately in the denominator?

That is a common trap! You cannot distribute the integral across a denominator. The expression $\frac{2^y}{2^y - 1}$ is a single functional unit. You must use substitution or algebraic manipulation to simplify the integrand before integrating.

Q: How did $\frac{1}{\ln 2}$ turn into $\log_2 e$?

Great observation! This is a standard logarithmic identity. Since $\log_b a = \frac{\ln a}{\ln b}$, it follows that $\frac{1}{\ln 2} = \log_2 e$. Using this identity makes the final expression much cleaner and easier to handle.

Q: Do I really need to find the constant $C$ before solving for $y(1)$?

Absolutely. The differential equation gives you a family of curves. The initial condition $y(0) = 1$ is the 'key' that selects the specific curve passing through that point. Without finding $C$, you cannot determine the unique value of $y(1)$.

Question

If $\frac{dy}{dx} = \frac{2^{x+y}-2^x}{2^y}, y(0) = 1$, then $y(1)$ is equal to :

Options:

(A) $\log_2(2 + e)$

(B) $\log_2(1 + e)$

(C) $\log_2(2e)$

(D) $\log_2(1 + e^2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex differential equation, and it feels like a tangled knot of exponents. The problem is:

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

We are given the initial condition $y(0) = 1$. While it looks intimidating, in the world of JEE Advanced, intimidation is often just a mask for elegance.

### The Art of Simplification

The first step is to recognize the structure. By the laws of exponents, we know that $2^{x+y} = 2^x \cdot 2^y$.

Substituting this back into the equation, we obtain:

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

Now, observe the numerator. We have a common factor of $2^x$. Factoring it out, we get:

$$\frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y}$$

Suddenly, the chaos has order. We have successfully isolated the $x$ and $y$ components.

### The Dance of Separation

Now, we use the **Variable Separable method**. We want all $y$ terms on the left and all $x$ terms on the right.

By cross-multiplying, we get:

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

This is the moment of truth. The variables are separated, and the path to integration is clear.

### The Calculus Core

We integrate both sides:

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

For the left-hand side, let $u = 2^y - 1$. Then $du = 2^y \ln 2 \, dy$, which implies $2^y dy = \frac{du}{\ln 2}$.

The integral becomes:

$$\frac{1}{\ln 2} \int \frac{1}{u} du = \frac{\ln(2^y - 1)}{\ln 2}$$

Using the change of base formula, this simplifies to $\log_2(2^y - 1)$. On the right-hand side, the integral of $2^x$ is $\frac{2^x}{\ln 2}$, which is $2^x \log_2 e$.

Adding our constant $C$, we get the general solution:

$$\log_2(2^y - 1) = 2^x \log_2 e + C$$

### The Final Reveal

We use the initial condition $y(0) = 1$ to find $C$. Substituting $x=0$ and $y=1$:

$$\log_2(2^1 - 1) = 2^0 \log_2 e + C$$

This simplifies to $\log_2(1) = \log_2 e + C$. Since $\log_2(1) = 0$, we find $C = -\log_2 e$.

Our particular solution is:

$$\log_2(2^y - 1) = (2^x - 1) \log_2 e$$

Finally, to find $y(1)$, we set $x=1$:

$$\log_2(2^y - 1) = (2^1 - 1) \log_2 e = \log_2 e$$

Comparing the arguments, $2^y - 1 = e$, so $2^y = 1 + e$. The final result is:

**$y = \log_2(1 + e)$**

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Select Answer:

Visualized Solution

Given Differential Equation

Simplifying the Expression

Variable Separable Method

Integrating Both Sides

LHS Integration

RHS Integration

General Solution

Finding the Constant

Evaluating C

Particular Solution

Finding y(1)

Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Simplification

The Dance of Separation

The Calculus Core

The Final Reveal

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :