Let y = y(x) be the solution curve of the differential equation dy/dx + 1/(x² - 1)y = ((x-1)/(x+1))^

Q: Why can't I just separate the variables? It looks like I could move $y$ to one side.

Great question! If you try to isolate $dy/y$, you will find that the $x$ terms are inextricably linked to the $y$ term via addition. This is a classic First-Order Linear Differential Equation. Variable separation only works if you can write the equation as $f(y)dy = g(x)dx$. Here, the additive term $\frac{1}{x^2-1}y$ prevents that. Always check for the linear form first!

Q: How do I simplify $e^{\frac{1}{2} \ln|\frac{x-1}{x+1}|}$?

This is a common stumbling block! Remember the logarithmic property $n \ln a = \ln(a^n)$. Move the $\frac{1}{2}$ inside the logarithm as an exponent. Since the exponent is $\frac{1}{2}$, it becomes a square root. Now you have $e^{\ln(\sqrt{\dots})}$. Because the exponential function and the natural logarithm are inverse functions, they effectively 'cancel' each other out, leaving you with just the argument of the log.

Q: Why do we need the initial condition $(2, \frac{1}{\sqrt{3}})$?

Differential equations describe a family of curves, not just one. The general solution contains an arbitrary constant $C$, which represents the vertical shifting of the curve. The initial condition acts as a 'pin' that locks the curve into a specific position on the coordinate plane. Without it, you would have an infinite number of solutions!

Question

Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \frac{1}{x^2 - 1}y = (\frac{x-1}{x+1})^{1/2}, x > 1$ passing through the point $(2, \frac{1}{\sqrt{3}})$. Then $\sqrt{7}y(8)$ is equal to

Options:

(A) $11 + 6\log_e 3$

(B) 19

(C) $12 - 2\log_e 3$

(D) $19 - 6\log_e 3$

Sigma Percentile · Accepted Answer

### Analyzing the Symphony of the Linear Differential Equation

Imagine standing before a complex differential equation:

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

In the world of JEE Advanced, intimidation is often just a mask for elegance. This problem is not a test of brute force; it is a test of pattern recognition.

### Phase 1: Recognizing the Pattern

The first step in any differential equation problem is to identify its soul. We see a derivative $\frac{dy}{dx}$, a term involving $y$, and a function of $x$.

This is the hallmark of a **First-Order Linear Differential Equation** of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

By identifying $P(x) = \frac{1}{x^2 - 1}$ and $Q(x) = \sqrt{\frac{x-1}{x+1}}$, we have already won half the battle. We have stripped away the complexity and revealed the underlying structure.

### Phase 2: The Integrating Factor

Now, we summon our most powerful tool: the Integrating Factor ($I.F.$). The formula is defined as:

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

Substituting our $P(x)$, we face the integral $\int \frac{1}{x^2 - 1} dx$. Using the standard integral formula $\frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right|$, with $a=1$, we obtain:

$$\int \frac{1}{x^2 - 1} dx = \frac{1}{2} \ln\left|\frac{x-1}{x+1}\right|$$

Here is where the magic happens. We use the logarithmic property $n \ln a = \ln(a^n)$ to bring the $\frac{1}{2}$ inside, turning it into a square root. Since $e^{\ln(z)} = z$, our $I.F.$ simplifies beautifully to:

$$I.F. = \sqrt{\frac{x-1}{x+1}}$$

### Phase 3: The Integration

With our $I.F.$ in hand, the general solution is given by:

$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

When we multiply $Q(x)$ by the $I.F.$, we get:

$$\int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} dx = \int \frac{x-1}{x+1} dx$$

To integrate this, we use a classic algebraic trick: rewrite the numerator as $(x+1) - 2$. This splits the integral into:

$$\int \left(1 - \frac{2}{x+1}\right) dx = x - 2\ln|x+1| + C$$

The terrifying integral has vanished, replaced by simple terms.

### Phase 4: The Final Victory

We are left with the general solution:

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

We use the given point $(2, \frac{1}{\sqrt{3}})$ to find the constant $C$. Substituting these values:

$$\frac{1}{\sqrt{3}} \cdot \sqrt{\frac{2-1}{2+1}} = 2 - 2\ln(3) + C \Rightarrow \frac{1}{3} = 2 - 2\ln 3 + C$$

Solving for $C$, we find $C = 2\ln 3 - \frac{5}{3}$. Finally, we evaluate the expression at $x=8$:

$$y(8) \cdot \sqrt{\frac{8-1}{8+1}} = 8 - 2\ln(9) + \left(2\ln 3 - \frac{5}{3}\right)$$

Since $\ln 9 = 2\ln 3$, the logarithmic terms combine to $-2\ln 3$, and the constants combine to $\frac{19}{3}$. Multiplying by $3$, we arrive at the elegant result:

$$\sqrt{7}y(8) = 19 - 6\ln 3$$

It was never about the complexity; it was about the journey of simplification.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x ^{2} - 1} y = (\frac{x - 1}{x + 1})^{1/2}, x > 1$ passing through the point $(2, \frac{1}{3})$ . Then $7 y (8)$ is equal to

Select Answer:

Visualized Solution

Standard Form of LDE

Identify $P (x)$ and $Q (x)$

Integrating Factor ( $I . F .$ ) Setup

Evaluate the Integral for $I . F .$

Simplify the $I . F .$ Expression

General Solution Equation

Simplify the Right Hand Side

Perform the Integration

Substitute Point $(2, \frac{1}{3})$

Calculate Constant $C$

Evaluate at $x = 8$

Final Calculation for $7 y (8)$

The Sigma Insight: Linear Differential Equations

Analyzing the Symphony of the Linear Differential Equation

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + lo g_{e} x)), x > 0, y (1) = 3$ . Then $\frac{y ^{2} ( x )}{9}$ is equal to :

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

If the curve $y = y (x)$ is the solution of the differential equation $2 (x^{2} + x^{5/4}) d y - y (x + x^{1/4}) d x = 2 x^{9/4} d x, x > 0$ which passes through the point $(1, 1 - \frac{4}{3} lo g_{e} 2)$ , then the value of $y (16)$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x ^{2} + 11 x + 13}{x ^{3} + 6 x ^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$ , which passes through the point $(0, 1)$ . Then $y (1)$ is equal to:

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:

If $y = y (x), x \in (0, \frac{π}{2})$ be the solution curve of the differential equation $(sin^{2} 2 x) \frac{d y}{d x} + (8 sin^{2} 2 x + 2 sin 4 x) y = 2 e^{- 4 x} (2 sin 2 x + cos 2 x)$ , with $y (\frac{π}{4}) = e^{- π}$ , then $y (\frac{π}{6})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y, y (1) = 0$ . Then $y (3)$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

Select Answer:

Visualized Solution

Standard Form of LDE

Identify .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)

Integrating Factor ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I.F.) Setup

Evaluate the Integral for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I.F.

Simplify the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I.F. Expression