2.4 Q-4

Question Statement

Prove that:
$y \frac{dy}{dx} + x = 0$
if:
$x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2}$

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Background and Explanation

To solve this, we need to:

1. Differentiate $x$ and $y$ with respect to $t$ to calculate $\frac{dy}{dx}$ using the chain rule.
2. Substitute the values of $x$, $y$, and $\frac{dy}{dx}$ into the given equation to verify it.

The chain rule states that:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Understanding parametric equations like $x(t)$ and $y(t)$ is necessary, as well as simplifying rational expressions.

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Solution

Step 1: Differentiate $y$ with respect to $t$

Given:
$y = \frac{2t}{1 + t^2}$
Differentiate with respect to $t$:

$$\frac{dy}{dt} = \frac{(1 + t^2)(2) - 2t(2t)}{(1 + t^2)^2}$$

Simplify the numerator:

$$\frac{dy}{dt} = \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2(1 - t^2)}{(1 + t^2)^2}$$

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Step 2: Differentiate $x$ with respect to $t$

Given:
$x = \frac{1 - t^2}{1 + t^2}$
Differentiate with respect to $t$:

$$\frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2}$$

Simplify the numerator:

$$\frac{dx}{dt} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}$$

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Step 3: Find $\frac{dy}{dx}$ using the chain rule

The chain rule gives:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Substitute the values of $\frac{dy}{dt}$ and $\frac{dx}{dt}$:

$$\frac{dy}{dx} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}}$$

Simplify:

$$\frac{dy}{dx} = \frac{2(1 - t^2)}{-4t} = \frac{1 - t^2}{-2t}$$

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Step 4: Verify the given equation

We need to show:

$$y \frac{dy}{dx} + x = 0$$

Substitute $y = \frac{2t}{1 + t^2}$ and $\frac{dy}{dx} = \frac{1 - t^2}{-2t}$:

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

Simplify:

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

Now, add $x = \frac{1 - t^2}{1 + t^2}$:

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

Combine the terms:

$$y \frac{dy}{dx} + x = \frac{1 - t^2 - (1 - t^2)}{1 + t^2} = \frac{0}{1 + t^2} = 0$$

Thus, the equation is verified.

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Key Formulas or Methods Used

1. Chain rule:
   $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
2. Differentiation of rational functions.
3. Simplifying expressions to verify the given equation.

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Summary of Steps

1. Differentiate $y$ with respect to $t$ to get $\frac{dy}{dt}$.
2. Differentiate $x$ with respect to $t$ to get $\frac{dx}{dt}$.
3. Use the chain rule to find $\frac{dy}{dx}$.
4. Substitute $x$, $y$, and $\frac{dy}{dx}$ into the equation $y \frac{dy}{dx} + x$ and simplify to verify it equals $0$.