Which of the following is true for y(x) that satisfies the differential equation dy/dx = xy - 1 + x

Q: Wait, why did we group the terms like that? How was I supposed to see that factorization?

Great question! In JEE problems, whenever you see a mix of $x$ and $y$ terms, always test for 'grouping'. If you see four terms, try pairing them up. Here, $xy - y$ and $x - 1$ share a common structure. It is a pattern-recognition skill that grows with practice!

Q: Can I just use the integrating factor method here?

You could, but it would be like using a sledgehammer to crack a nut! The integrating factor method is for linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Since our equation is separable, that method is unnecessary and much more prone to calculation errors.

Q: Why did the absolute value sign in $\ln|y+1|$ disappear?

Excellent observation. We use the initial condition $y(0)=0$. At $x=0$, $y+1 = 1$, which is positive. Since the function is continuous, $y+1$ remains positive in the neighborhood of our interest, allowing us to drop the absolute value bars safely.

Q: I got $C=0$, but I am always worried about forgetting the constant. Is it really that simple?

It is! Never skip the constant $C$ during indefinite integration. Even if it turns out to be zero, writing it down ensures you don't lose points in subjective steps or make a mistake in more complex problems where $C$ is non-zero.

Q: Is there any other way to solve this if I don't see the factorization?

If you absolutely cannot see the factorization, you could try to treat it as a Bernoulli equation or check for exactness, but those are significantly longer paths. In competitive exams, the 'hidden' factorization is almost always the intended, elegant 'shortcut' designed by the examiner.

Question

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

Options:

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

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

(C) $y(1) = 1$

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

Sigma Percentile · Accepted Answer

### The Art of Seeing the Hidden Structure

Welcome, fellow traveler on the road to JEE excellence. Today, we are going to dismantle a differential equation that, at first glance, looks like a tangled mess of variables.

You might look at $\frac{dy}{dx} = xy - 1 + x - y$ and feel a sense of dread. But I want you to take a deep breath. In mathematics, as in life, complexity is often just simplicity in disguise.

Our goal is to peel back the layers of this equation until the underlying structure reveals itself.

### Phase 1: The Power of Grouping

When you encounter a differential equation where $x$ and $y$ are intertwined, your first instinct should be to look for symmetry. Look at the right-hand side: $xy - 1 + x - y$.

If we rearrange these terms, we get $(xy + x) - (y + 1)$. By grouping the terms, we can pull out a common factor of $x$ from the first pair and a factor of $1$ from the second.

This transforms our equation into:

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

Suddenly, the fog clears. We have a common binomial factor of $(y + 1)$. Factoring this out, we arrive at the elegant form:

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

This is the **'Aha!' moment**. We have successfully separated the variables. We have turned a chaotic expression into a product of a function of $y$ and a function of $x$.

### Phase 2: The Dance of Integration

Now that we have $\frac{dy}{dx} = (y + 1)(x - 1)$, we can move all the $y$ terms to one side and all the $x$ terms to the other. This is the heart of the **Variable Separable** method:

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

Integrating both sides is our next logical step. We are looking for the antiderivative of $\frac{1}{y+1}$ with respect to $y$, and $(x-1)$ with respect to $x$.

This yields:

$$\int \frac{dy}{y + 1} = \int (x - 1) dx$$

Performing the integration, we get:

$$\ln|y + 1| = \frac{x^2}{2} - x + C$$

### Phase 3: The Initial Condition and the Final Reveal

We are almost there. We have a general solution, but we need the particular solution that satisfies $y(0) = 0$.

By substituting $x = 0$ and $y = 0$ into our equation, we find:

$$\ln|0 + 1| = \frac{0^2}{2} - 0 + C \Rightarrow \ln(1) = 0 + C \Rightarrow C = 0$$

With $C$ identified as zero, our equation simplifies beautifully to $\ln(y + 1) = \frac{x^2}{2} - x$. To isolate $y$, we exponentiate both sides:

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

Finally, to find $y(1)$, we simply plug in $x = 1$:

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

Look at that! Through patience and systematic grouping, we have navigated from a daunting differential equation to a precise, elegant result.

The final answer is **$y(1) = e^{-1/2} - 1$**. Remember, the next time you face a problem that looks impossible, don't panic. Look for the hidden factors, trust your process, and let the math lead you home.

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

Select Answer:

Visualized Solution

Analyze the Differential Equation

Factorizing the Expression

Final Factored Form

Variable Separation

Integrating Both Sides

Executing Integration

Applying Initial Condition

Finding the Constant $C$

The Particular Solution

Explicit Form of Solution

Calculating $y (1)$ - Setup

Calculating $y (1)$ - Execution

Final Conclusion

The Sigma Insight: Variable Separable Method

The Art of Seeing the Hidden Structure

Phase 1

Phase 2

Phase 3

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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; } Which of the following is true for y(x) that satisfies the differential equation dxdy​=xy−1+x−y;y(0)=0