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Given x = (1−t2)/(1+t2).

Mathematics

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Answer

dydx+xy=0\frac{dy}{dx} + \frac{x}{y} = 0

Step-by-step solution

5 steps

  1. Find dxdt\frac{dx}{dt} using the quotient rule.

    Given x=1−t21+t2x = \frac{1-t^2}{1+t^2}.
    Let u=1−t2u = 1-t^2 and v=1+t2v = 1+t^2. Then dudt=−2t\frac{du}{dt} = -2t and dvdt=2t\frac{dv}{dt} = 2t.
    The quotient rule is ddt(uv)=u′v−uv′v2\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

    dxdt=(−2t)(1+t2)−(1−t2)(2t)(1+t2)2\frac{dx}{dt} = \frac{(-2t)(1+t^2) - (1-t^2)(2t)}{(1+t^2)^2} dxdt=−2t−2t3−(2t−2t3)(1+t2)2\frac{dx}{dt} = \frac{-2t - 2t^3 - (2t - 2t^3)}{(1+t^2)^2} dxdt=−2t−2t3−2t+2t3(1+t2)2\frac{dx}{dt} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1+t^2)^2} dxdt=−4t(1+t2)2\frac{dx}{dt} = \frac{-4t}{(1+t^2)^2}

  2. Find dydt\frac{dy}{dt} using the quotient rule.

    Given y=2t1+t2y = \frac{2t}{1+t^2}.
    Let u=2tu = 2t and v=1+t2v = 1+t^2. Then dudt=2\frac{du}{dt} = 2 and dvdt=2t\frac{dv}{dt} = 2t.

    dydt=(2)(1+t2)−(2t)(2t)(1+t2)2\frac{dy}{dt} = \frac{(2)(1+t^2) - (2t)(2t)}{(1+t^2)^2} dydt=2+2t2−4t2(1+t2)2\frac{dy}{dt} = \frac{2 + 2t^2 - 4t^2}{(1+t^2)^2} dydt=2−2t2(1+t2)2\frac{dy}{dt} = \frac{2 - 2t^2}{(1+t^2)^2} dydt=2(1−t2)(1+t2)2\frac{dy}{dt} = \frac{2(1 - t^2)}{(1+t^2)^2}

  3. Calculate dydx\frac{dy}{dx} using the chain rule dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

    dydx=2(1−t2)(1+t2)2−4t(1+t2)2\frac{dy}{dx} = \frac{\frac{2(1 - t^2)}{(1+t^2)^2}}{\frac{-4t}{(1+t^2)^2}} dydx=2(1−t2)(1+t2)2⋅(1+t2)2−4t\frac{dy}{dx} = \frac{2(1 - t^2)}{(1+t^2)^2} \cdot \frac{(1+t^2)^2}{-4t} dydx=2(1−t2)−4t\frac{dy}{dx} = \frac{2(1 - t^2)}{-4t} dydx=−1−t22t\frac{dy}{dx} = -\frac{1 - t^2}{2t}

  4. Calculate xy\frac{x}{y}.

    x=1−t21+t2x = \frac{1-t^2}{1+t^2} y=2t1+t2y = \frac{2t}{1+t^2} xy=1−t21+t22t1+t2\frac{x}{y} = \frac{\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} xy=1−t21+t2⋅1+t22t\frac{x}{y} = \frac{1-t^2}{1+t^2} \cdot \frac{1+t^2}{2t} xy=1−t22t\frac{x}{y} = \frac{1-t^2}{2t}

  5. Substitute dydx\frac{dy}{dx} and xy\frac{x}{y} into the expression dydx+xy\frac{dy}{dx} + \frac{x}{y}.

    dydx+xy=−1−t22t+1−t22t\frac{dy}{dx} + \frac{x}{y} = -\frac{1 - t^2}{2t} + \frac{1-t^2}{2t} dydx+xy=0\frac{dy}{dx} + \frac{x}{y} = 0
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