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EXERCISE 1.5 · Q190

Q.x=at2x=at^2, y=2aty=2at, then show that xy d2ydx2+a=0xy\,\dfrac{d^2y}{dx^2}+a=0

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Given x=at2x=at^2, y=2aty=2at.

Step 1: dxdt=2at\dfrac{dx}{dt}=2at, dydt=2a\dfrac{dy}{dt}=2a, so y1=dydx=2a2at=1ty_1=\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac1t.

Step 2: ddt(y1)=−1t2\dfrac{d}{dt}(y_1)=-\dfrac{1}{t^2}, and dividing by dx/dt=2atdx/dt=2at gives y2=d2ydx2=−1/t22at=−12at3y_2=\dfrac{d^2y}{dx^2}=\dfrac{-1/t^2}{2at}=-\dfrac{1}{2at^3}.

Step 3 — compute xyxy: xy=at2⋅2at=2a2t3xy=at^2\cdot2at=2a^2t^3. …

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