If x³dy + xydx = x²dy + 2ydx; y(2) = e and x 1, then y(4) is equal to : (A) √e/2 (B) 1/2 + √e (C)

Q: Wait, why can't I use the integrating factor method ($e^{\int P dx}$) here?

That is a fantastic question! The integrating factor method is a powerful tool, but it is exclusively reserved for linear differential equations of the form $\frac{dy}{dx} + Py = Q$. If you look at our equation, the $y$ terms are multiplied by $x$ in a way that doesn't fit that linear structure. Always check the structure of your equation before choosing your weapon. In this case, the variables were separable, which is a much more direct path.

Q: How do I know if I need partial fractions? It looks like a lot of extra work.

Partial fractions are the surgical tools of calculus. Whenever you see a rational function—a polynomial divided by another polynomial—where the denominator is factorable, partial fractions are your best friend. It turns a nightmare integral into a sum of simple, standard integrals. It might feel like extra work, but it is the most reliable way to break down complex expressions into manageable pieces.

Q: Why did the constant $K$ turn out to be $\ln 2$? Did I do something wrong?

Not at all! You did perfectly. Remember that the constant of integration $K$ is just a placeholder for a real number. When we use the initial condition $y(2) = e$, we are essentially solving for the specific 'DNA' of this curve. Since our equation involves natural logarithms, it is mathematically elegant to express the constant as a logarithm as well. It makes the final simplification much cleaner.

Question

If $x^3dy + xydx = x^2dy + 2ydx$; $y(2) = e$ and $x > 1$, then $y(4)$ is equal to :

Options:

(A) $\frac{\sqrt{e}}{2}$

(B) $\frac{1}{2} + \sqrt{e}$

(C) $\frac{3}{2} + \sqrt{e}$

(D) $\frac{3}{2}\sqrt{e}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to peel back the layers of a differential equation that might look like a chaotic jumble of symbols at first glance. But remember, in the world of JEE Advanced, chaos is just order waiting to be discovered.

We are looking at the equation:
$$x^3 dy + xy dx = x^2 dy + 2y dx$$

Our mission is to find the function $y(x)$ that satisfies this relationship, given the initial condition $y(2) = e$. Let us embark on this journey.

### The Art of Separation

The first step in any differential equation is to impose order. We have $dy$ and $dx$ terms scattered across both sides of the equation. Our goal is to group them.

By moving the $dy$ terms to the left and the $dx$ terms to the right, we transform the equation into:
$$x^3 dy - x^2 dy = 2y dx - xy dx$$

Now, we factor out the common terms. On the left, we pull out $x^2 dy$, and on the right, we pull out $y dx$. This gives us:
$$x^2(x - 1) dy = y(2 - x) dx$$

Suddenly, the fog clears. We can now separate the variables completely:
$$\frac{dy}{y} = \frac{2 - x}{x^2(x - 1)} dx$$

This is the beauty of variable separation—we have effectively isolated the influence of $y$ from the influence of $x$.

### The Surgical Precision of Partial Fractions

Now, we integrate both sides. The left side is trivial: the integral of $\frac{1}{y} dy$ is simply $\ln|y|$.

But the right side requires the integral of a rational function:
$$\int \frac{2 - x}{x^2(x - 1)} dx$$

We cannot integrate this directly. We must use partial fraction decomposition. We express the integrand as:
$$\frac{2 - x}{x^2(x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}$$

By multiplying through by the denominator, we get the master equation:
$$2 - x = Ax(x - 1) + B(x - 1) + Cx^2$$

This is where we play the game of coefficients. By setting $x=0$, we find $B = -2$. By setting $x=1$, we find $C = 1$. Finally, by comparing the coefficients of $x^2$, we find $A = -1$.

### The Final Synthesis

With our constants $A, B,$ and $C$ in hand, our integral becomes:
$$\int \left( -\frac{1}{x} - \frac{2}{x^2} + \frac{1}{x - 1} \right) dx$$

Integrating term by term, we get:
$$\ln|y| = -\ln|x| + \frac{2}{x} + \ln|x - 1| + K$$

Using the properties of logarithms, we can condense this into:
$$\ln\left| \frac{y \cdot x}{x - 1} \right| = \frac{2}{x} + K$$

Now, we apply our initial condition $y(2) = e$. Substituting $x=2$ and $y=e$, we find that $K = \ln 2$.

Our particular solution is now complete:
$$\ln\left| \frac{yx}{x - 1} \right| = \frac{2}{x} + \ln 2$$

Finally, to find $y(4)$, we substitute $x=4$ into our equation. After a bit of algebraic cleanup, we arrive at the elegant result:
**$$y = \frac{3}{2}e^{1/2}$$**

You have navigated the chaos, applied the tools, and arrived at the truth. This is the essence of mathematics—finding the hidden order in the complexity.

Animated Solution for Mathematics - Differential Equations: If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Grouping $d y$ and $d x$ Terms

Separating the Variables

Integrating Both Sides

Partial Fraction Decomposition

Solving for Coefficients $B$ and $C$

Solving for Coefficient $A$

Integrating the RHS Terms

Simplifying with Log Properties

Applying the Initial Condition

Finding the Constant $K$

Substituting $x = 4$

Final Answer Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Surgical Precision of Partial Fractions

The Final Synthesis

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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 x3dy+xydx=x2dy+2ydx; y(2)=e and x>1, then y(4) is equal to :

Select Answer:

Visualized Solution

Analyze the Differential Equation

Grouping .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dy and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx Terms

Separating the Variables

Integrating Both Sides

Partial Fraction Decomposition

Solving for Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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