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2.4 Q-4

Question Statement

Prove that:
ydydx+x=0y \frac{dy}{dx} + x = 0
if:
x=1βˆ’t21+t2,y=2t1+t2x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2}


Background and Explanation

To solve this, we need to:

  1. Differentiate xx and yy with respect to tt to calculate dydx\frac{dy}{dx} using the chain rule.
  2. Substitute the values of xx, yy, and dydx\frac{dy}{dx} into the given equation to verify it.

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

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


Solution

Step 1: Differentiate yy with respect to tt

Given:
y=2t1+t2y = \frac{2t}{1 + t^2}
Differentiate with respect to tt:

dydt=(1+t2)(2)βˆ’2t(2t)(1+t2)2\frac{dy}{dt} = \frac{(1 + t^2)(2) - 2t(2t)}{(1 + t^2)^2}

Simplify the numerator:

dydt=2+2t2βˆ’4t2(1+t2)2=2(1βˆ’t2)(1+t2)2\frac{dy}{dt} = \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2(1 - t^2)}{(1 + t^2)^2}

Step 2: Differentiate xx with respect to tt

Given:
x=1βˆ’t21+t2x = \frac{1 - t^2}{1 + t^2}
Differentiate with respect to tt:

dxdt=(1+t2)(βˆ’2t)βˆ’(1βˆ’t2)(2t)(1+t2)2\frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2}

Simplify the numerator:

dxdt=βˆ’2tβˆ’2t3βˆ’2t+2t3(1+t2)2=βˆ’4t(1+t2)2\frac{dx}{dt} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}

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

The chain rule gives:

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

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

dydx=2(1βˆ’t2)(1+t2)2βˆ’4t(1+t2)2\frac{dy}{dx} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}}

Simplify:

dydx=2(1βˆ’t2)βˆ’4t=1βˆ’t2βˆ’2t\frac{dy}{dx} = \frac{2(1 - t^2)}{-4t} = \frac{1 - t^2}{-2t}

Step 4: Verify the given equation

We need to show:

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

Substitute y=2t1+t2y = \frac{2t}{1 + t^2} and dydx=1βˆ’t2βˆ’2t\frac{dy}{dx} = \frac{1 - t^2}{-2t}:

ydydx=(2t1+t2)(1βˆ’t2βˆ’2t)y \frac{dy}{dx} = \left(\frac{2t}{1 + t^2}\right)\left(\frac{1 - t^2}{-2t}\right)

Simplify:

ydydx=(1βˆ’t2)βˆ’(1+t2)y \frac{dy}{dx} = \frac{(1 - t^2)}{-(1 + t^2)}

Now, add x=1βˆ’t21+t2x = \frac{1 - t^2}{1 + t^2}:

ydydx+x=(1βˆ’t2)βˆ’(1+t2)+1βˆ’t21+t2y \frac{dy}{dx} + x = \frac{(1 - t^2)}{-(1 + t^2)} + \frac{1 - t^2}{1 + t^2}

Combine the terms:

ydydx+x=1βˆ’t2βˆ’(1βˆ’t2)1+t2=01+t2=0y \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.


Key Formulas or Methods Used

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

Summary of Steps

  1. Differentiate yy with respect to tt to get dydt\frac{dy}{dt}.
  2. Differentiate xx with respect to tt to get dxdt\frac{dx}{dt}.
  3. Use the chain rule to find dydx\frac{dy}{dx}.
  4. Substitute xx, yy, and dydx\frac{dy}{dx} into the equation ydydx+xy \frac{dy}{dx} + x and simplify to verify it equals 00.