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EXERCISE 1.5 · Q201

Q.x=asin⁡t−bcos⁡tx=a\sin t-b\cos t, y=acos⁡t+bsin⁡ty=a\cos t+b\sin t, show that d2ydx2=−x2+y2y3\dfrac{d^2y}{dx^2}=-\dfrac{x^2+y^2}{y^3}

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Given x=asin⁡t−bcos⁡tx=a\sin t-b\cos t, y=acos⁡t+bsin⁡ty=a\cos t+b\sin t.

Step 1 — differentiate: dxdt=acos⁡t+bsin⁡t=y\dfrac{dx}{dt}=a\cos t+b\sin t=y and dydt=−asin⁡t+bcos⁡t=−(asin⁡t−bcos⁡t)=−x\dfrac{dy}{dt}=-a\sin t+b\cos t=-(a\sin t-b\cos t)=-x.

Step 2 — so y1=dydx=dy/dtdx/dt=−xyy_1=\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{-x}{y}. …

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