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Question: If $2x = (y^{\frac{1}{3}} + y^{-\frac{1}{3}})$, then the value of $\frac{(x^2 - 1)}{y} \cdot \frac{d...

If 2x=(y13+y−13)2x = (y^{\frac{1}{3}} + y^{-\frac{1}{3}}), then the value of (x2−1)y⋅d2ydx2+xy⋅dydx\frac{(x^2 - 1)}{y} \cdot \frac{d^2y}{dx^2} + \frac{x}{y} \cdot \frac{dy}{dx}, is

Answer

9

Explanation

Solution

Given the relation 2x=y13+y−132x = y^{\frac{1}{3}} + y^{-\frac{1}{3}}. We want to find the value of the expression E=(x2−1)y⋅d2ydx2+xy⋅dydxE = \frac{(x^2 - 1)}{y} \cdot \frac{d^2y}{dx^2} + \frac{x}{y} \cdot \frac{dy}{dx}.

First, let's find dydx\frac{dy}{dx}. Differentiate the given relation with respect to xx: ddx(2x)=ddx(y1/3+y−1/3)\frac{d}{dx}(2x) = \frac{d}{dx}(y^{1/3} + y^{-1/3}) 2=13y1/3−1dydx−13y−1/3−1dydx2 = \frac{1}{3} y^{1/3 - 1} \frac{dy}{dx} - \frac{1}{3} y^{-1/3 - 1} \frac{dy}{dx} 2=13y−2/3dydx−13y−4/3dydx2 = \frac{1}{3} y^{-2/3} \frac{dy}{dx} - \frac{1}{3} y^{-4/3} \frac{dy}{dx} 2=13(y−2/3−y−4/3)dydx2 = \frac{1}{3} (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} 6=(y−2/3−y−4/3)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} 6=y2/3−1y4/3dydx6 = \frac{y^{2/3} - 1}{y^{4/3}} \frac{dy}{dx} dydx=6y4/3y2/3−1\frac{dy}{dx} = \frac{6 y^{4/3}}{y^{2/3} - 1}.

Now let's find d2ydx2\frac{d^2y}{dx^2}. Differentiate the equation 6=(y−2/3−y−4/3)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} with respect to xx: ddx(6)=ddx((y−2/3−y−4/3)dydx)\frac{d}{dx}(6) = \frac{d}{dx}\left((y^{-2/3} - y^{-4/3}) \frac{dy}{dx}\right) 0=(ddx(y−2/3−y−4/3))dydx+(y−2/3−y−4/3)d2ydx20 = \left(\frac{d}{dx}(y^{-2/3} - y^{-4/3})\right) \frac{dy}{dx} + (y^{-2/3} - y^{-4/3}) \frac{d^2y}{dx^2} 0=(−23y−5/3dydx+43y−7/3dydx)dydx+(y−2/3−y−4/3)d2ydx20 = \left(-\frac{2}{3} y^{-5/3} \frac{dy}{dx} + \frac{4}{3} y^{-7/3} \frac{dy}{dx}\right) \frac{dy}{dx} + (y^{-2/3} - y^{-4/3}) \frac{d^2y}{dx^2} 0=23y−7/3(2−y2/3)(dydx)2+y−4/3(y2/3−1)d2ydx20 = \frac{2}{3} y^{-7/3} (2 - y^{2/3}) \left(\frac{dy}{dx}\right)^2 + y^{-4/3} (y^{2/3} - 1) \frac{d^2y}{dx^2}. Multiply by y4/3y^{4/3}: 0=23y−1(2−y2/3)(dydx)2+(y2/3−1)d2ydx20 = \frac{2}{3} y^{-1} (2 - y^{2/3}) \left(\frac{dy}{dx}\right)^2 + (y^{2/3} - 1) \frac{d^2y}{dx^2} (y2/3−1)d2ydx2=−23y(2−y2/3)(dydx)2(y^{2/3} - 1) \frac{d^2y}{dx^2} = - \frac{2}{3y} (2 - y^{2/3}) \left(\frac{dy}{dx}\right)^2 (y2/3−1)d2ydx2=23y(y2/3−2)(dydx)2(y^{2/3} - 1) \frac{d^2y}{dx^2} = \frac{2}{3y} (y^{2/3} - 2) \left(\frac{dy}{dx}\right)^2.

Consider the expression to be evaluated: E=(x2−1)yd2ydx2+xydydxE = \frac{(x^2 - 1)}{y} \frac{d^2y}{dx^2} + \frac{x}{y} \frac{dy}{dx}. Let's rearrange the equation for d2ydx2\frac{d^2y}{dx^2}: d2ydx2=2(y2/3−2)3y(y2/3−1)(dydx)2\frac{d^2y}{dx^2} = \frac{2(y^{2/3} - 2)}{3y(y^{2/3} - 1)} \left(\frac{dy}{dx}\right)^2. Substitute this into the expression for EE: E=(x2−1)y(2(y2/3−2)3y(y2/3−1)(dydx)2)+xydydxE = \frac{(x^2 - 1)}{y} \left( \frac{2(y^{2/3} - 2)}{3y(y^{2/3} - 1)} \left(\frac{dy}{dx}\right)^2 \right) + \frac{x}{y} \frac{dy}{dx} E=2(x2−1)(y2/3−2)3y2(y2/3−1)(dydx)2+xydydxE = \frac{2(x^2 - 1)(y^{2/3} - 2)}{3y^2(y^{2/3} - 1)} \left(\frac{dy}{dx}\right)^2 + \frac{x}{y} \frac{dy}{dx}.

This approach seems complicated due to the powers of yy. Let's try expressing x2−1x^2 - 1 in terms of yy. 2x=y1/3+y−1/32x = y^{1/3} + y^{-1/3} 4x2=(y1/3+y−1/3)2=y2/3+y−2/3+24x^2 = (y^{1/3} + y^{-1/3})^2 = y^{2/3} + y^{-2/3} + 2 4x2−4=y2/3+y−2/3−2=(y1/3−y−1/3)24x^2 - 4 = y^{2/3} + y^{-2/3} - 2 = (y^{1/3} - y^{-1/3})^2 x2−1=14(y1/3−y−1/3)2x^2 - 1 = \frac{1}{4} (y^{1/3} - y^{-1/3})^2.

Let's rewrite the equation 6=(y−2/3−y−4/3)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} as 6y4/3=(y2/3−1)dydx6 y^{4/3} = (y^{2/3} - 1) \frac{dy}{dx}. Differentiating this with respect to xx: ddx(6y4/3)=ddx((y2/3−1)dydx)\frac{d}{dx}(6 y^{4/3}) = \frac{d}{dx}((y^{2/3} - 1) \frac{dy}{dx}) 6⋅43y1/3dydx=(ddx(y2/3−1))dydx+(y2/3−1)d2ydx26 \cdot \frac{4}{3} y^{1/3} \frac{dy}{dx} = \left(\frac{d}{dx}(y^{2/3} - 1)\right) \frac{dy}{dx} + (y^{2/3} - 1) \frac{d^2y}{dx^2} 8y1/3dydx=23y−1/3dydx⋅dydx+(y2/3−1)d2ydx28 y^{1/3} \frac{dy}{dx} = \frac{2}{3} y^{-1/3} \frac{dy}{dx} \cdot \frac{dy}{dx} + (y^{2/3} - 1) \frac{d^2y}{dx^2} 8y1/3dydx=23y−1/3(dydx)2+(y2/3−1)d2ydx28 y^{1/3} \frac{dy}{dx} = \frac{2}{3} y^{-1/3} \left(\frac{dy}{dx}\right)^2 + (y^{2/3} - 1) \frac{d^2y}{dx^2}. Multiply by y1/3y^{1/3}: 8y2/3dydx=23(dydx)2+y(y2/3−1)d2ydx28 y^{2/3} \frac{dy}{dx} = \frac{2}{3} \left(\frac{dy}{dx}\right)^2 + y (y^{2/3} - 1) \frac{d^2y}{dx^2}. Multiply by 32\frac{3}{2}: 12y2/3dydx=(dydx)2+32y(y2/3−1)d2ydx212 y^{2/3} \frac{dy}{dx} = \left(\frac{dy}{dx}\right)^2 + \frac{3}{2} y (y^{2/3} - 1) \frac{d^2y}{dx^2}.

This still involves powers of yy. Let's try to use the relation x2−1=14(y1/3−y−1/3)2x^2 - 1 = \frac{1}{4}(y^{1/3} - y^{-1/3})^2. From 6=(y−2/3−y−4/3)dydx=y−4/3(y2/3−1)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} = y^{-4/3} (y^{2/3} - 1) \frac{dy}{dx}. dydx=6y4/3y2/3−1\frac{dy}{dx} = \frac{6 y^{4/3}}{y^{2/3} - 1}. dydx=6y(y2/3−1)y−1/3\frac{dy}{dx} = \frac{6 y}{(y^{2/3} - 1) y^{-1/3}}.

Consider the expression E=(x2−1)yd2ydx2+xydydxE = \frac{(x^2 - 1)}{y} \frac{d^2y}{dx^2} + \frac{x}{y} \frac{dy}{dx}. Multiply by yy: yE=(x2−1)d2ydx2+xdydxyE = (x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx}. This is in the form of the derivative of a product. Let's check if it is related to the derivative of (x2−1)dydx(x^2 - 1) \frac{dy}{dx}. ddx((x2−1)dydx)=2xdydx+(x2−1)d2ydx2\frac{d}{dx} \left( (x^2 - 1) \frac{dy}{dx} \right) = 2x \frac{dy}{dx} + (x^2 - 1) \frac{d^2y}{dx^2}. This is close, but the 2x2x term is different.

Let's consider the derivative of xdydxx \frac{dy}{dx}: ddx(xdydx)=dxdxdydx+xd2ydx2=dydx+xd2ydx2\frac{d}{dx} \left( x \frac{dy}{dx} \right) = \frac{dx}{dx} \frac{dy}{dx} + x \frac{d^2y}{dx^2} = \frac{dy}{dx} + x \frac{d^2y}{dx^2}.

Let's consider the derivative of (x2−1)dydx(x^2-1) \frac{dy}{dx} again. ddx((x2−1)dydx)=2xdydx+(x2−1)d2ydx2\frac{d}{dx} \left( (x^2-1) \frac{dy}{dx} \right) = 2x \frac{dy}{dx} + (x^2-1) \frac{d^2y}{dx^2}.

The expression is (x2−1)yd2ydx2+xydydx=1y[(x2−1)d2ydx2+xdydx]\frac{(x^2 - 1)}{y} \frac{d^2y}{dx^2} + \frac{x}{y} \frac{dy}{dx} = \frac{1}{y} \left[ (x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} \right]. Let's look at the structure of the derivatives we found. 6=(y−2/3−y−4/3)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx}. 0=23y−7/3(2−y2/3)(dydx)2+y−4/3(y2/3−1)d2ydx20 = \frac{2}{3} y^{-7/3} (2 - y^{2/3}) \left(\frac{dy}{dx}\right)^2 + y^{-4/3} (y^{2/3} - 1) \frac{d^2y}{dx^2}.

Let's consider the derivative of dydx\frac{dy}{dx} with respect to yy. dxdy=12(13y−2/3−13y−4/3)=16(y−2/3−y−4/3)\frac{dx}{dy} = \frac{1}{2} (\frac{1}{3} y^{-2/3} - \frac{1}{3} y^{-4/3}) = \frac{1}{6} (y^{-2/3} - y^{-4/3}). dydx=1dxdy=6y−2/3−y−4/3=6y4/3y2/3−1\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{6}{y^{-2/3} - y^{-4/3}} = \frac{6 y^{4/3}}{y^{2/3} - 1}. This matches.

Now find d2ydx2=ddx(dydx)=ddy(dydx)dydx\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dy} \left(\frac{dy}{dx}\right) \frac{dy}{dx}. ddy(6y4/3y2/3−1)=643y1/3(y2/3−1)−y4/3(23y−1/3)(y2/3−1)2\frac{d}{dy} \left(\frac{6 y^{4/3}}{y^{2/3} - 1}\right) = 6 \frac{\frac{4}{3} y^{1/3} (y^{2/3} - 1) - y^{4/3} (\frac{2}{3} y^{-1/3})}{(y^{2/3} - 1)^2} =643y−43y1/3−23y(y2/3−1)2=623y−43y1/3(y2/3−1)2=4y−2y1/3(y2/3−1)2=4y1/3y2/3−2(y2/3−1)2= 6 \frac{\frac{4}{3} y - \frac{4}{3} y^{1/3} - \frac{2}{3} y}{(y^{2/3} - 1)^2} = 6 \frac{\frac{2}{3} y - \frac{4}{3} y^{1/3}}{(y^{2/3} - 1)^2} = 4 \frac{y - 2 y^{1/3}}{(y^{2/3} - 1)^2} = 4 y^{1/3} \frac{y^{2/3} - 2}{(y^{2/3} - 1)^2}. d2ydx2=4y1/3y2/3−2(y2/3−1)2⋅6y4/3y2/3−1=24y5/3(y2/3−2)(y2/3−1)3\frac{d^2y}{dx^2} = 4 y^{1/3} \frac{y^{2/3} - 2}{(y^{2/3} - 1)^2} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} = \frac{24 y^{5/3} (y^{2/3} - 2)}{(y^{2/3} - 1)^3}. This also matches.

Let's use the relation x2−1=14(y1/3−y−1/3)2=14(y2/3−1)2y2/3x^2 - 1 = \frac{1}{4}(y^{1/3} - y^{-1/3})^2 = \frac{1}{4} \frac{(y^{2/3} - 1)^2}{y^{2/3}}. Substitute this into the expression for EE. E=14(y2/3−1)2y2/3y⋅24y5/3(y2/3−2)(y2/3−1)3+12(y1/3+y−1/3)y⋅6y4/3y2/3−1E = \frac{\frac{1}{4} \frac{(y^{2/3} - 1)^2}{y^{2/3}}}{y} \cdot \frac{24 y^{5/3} (y^{2/3} - 2)}{(y^{2/3} - 1)^3} + \frac{\frac{1}{2}(y^{1/3} + y^{-1/3})}{y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=14y5/3(y2/3−1)2(y2/3−1)3⋅24y5/3(y2/3−2)+y1/3+y−1/32y⋅6y4/3y2/3−1E = \frac{1}{4 y^{5/3}} \frac{(y^{2/3} - 1)^2}{(y^{2/3} - 1)^3} \cdot 24 y^{5/3} (y^{2/3} - 2) + \frac{y^{1/3} + y^{-1/3}}{2y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=14y5/31y2/3−1⋅24y5/3(y2/3−2)+y1/3+y−1/32y⋅6y4/3y2/3−1E = \frac{1}{4 y^{5/3}} \frac{1}{y^{2/3} - 1} \cdot 24 y^{5/3} (y^{2/3} - 2) + \frac{y^{1/3} + y^{-1/3}}{2y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=6y2/3−2y2/3−1+y1/3(1+y−2/3)2y⋅6y4/3y2/3−1E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{y^{1/3}(1 + y^{-2/3})}{2y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=6y2/3−2y2/3−1+y1/3y2/3+1y2/32y⋅6y4/3y2/3−1E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{y^{1/3} \frac{y^{2/3} + 1}{y^{2/3}}}{2y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=6y2/3−2y2/3−1+y2/3+12y5/3⋅6y4/3y2/3−1E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{y^{2/3} + 1}{2 y^{5/3}} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} E=6y2/3−2y2/3−1+6y4/3(y2/3+1)2y5/3(y2/3−1)E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{6 y^{4/3} (y^{2/3} + 1)}{2 y^{5/3} (y^{2/3} - 1)} E=6y2/3−2y2/3−1+3y−1/3(y2/3+1)y2/3−1=6y2/3−2y2/3−1+3(1+y−2/3)y2/3−1E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{3 y^{-1/3} (y^{2/3} + 1)}{y^{2/3} - 1} = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + \frac{3 (1 + y^{-2/3})}{y^{2/3} - 1}. This is not matching the previous calculation. Let's recheck the second term substitution.

Second term: xy⋅dydx\frac{x}{y} \cdot \frac{dy}{dx} x=12(y1/3+y−1/3)x = \frac{1}{2}(y^{1/3} + y^{-1/3}) xy=y1/3+y−1/32y\frac{x}{y} = \frac{y^{1/3} + y^{-1/3}}{2y} dydx=6y4/3y2/3−1\frac{dy}{dx} = \frac{6 y^{4/3}}{y^{2/3} - 1} xy⋅dydx=y1/3+y−1/32y⋅6y4/3y2/3−1=(y1/3+y−1/3)3y4/3y(y2/3−1)\frac{x}{y} \cdot \frac{dy}{dx} = \frac{y^{1/3} + y^{-1/3}}{2y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} = \frac{(y^{1/3} + y^{-1/3}) 3 y^{4/3}}{y(y^{2/3} - 1)} =(y1/3+y−1/3)3y4/3y5/3(y2/3−1)=3(y5/3+y1/3)y5/3(y2/3−1)=3y1/3(y4/3+1)y5/3(y2/3−1)=3(y4/3+1)y4/3(y2/3−1)= \frac{(y^{1/3} + y^{-1/3}) 3 y^{4/3}}{y^{5/3} (y^{2/3} - 1)} = \frac{3 (y^{5/3} + y^{1/3})}{y^{5/3} (y^{2/3} - 1)} = \frac{3 y^{1/3} (y^{4/3} + 1)}{y^{5/3} (y^{2/3} - 1)} = \frac{3 (y^{4/3} + 1)}{y^{4/3} (y^{2/3} - 1)}. This is also not matching.

Let's recheck the second term calculation in the thought block. Second term: xy⋅dydx\frac{x}{y} \cdot \frac{dy}{dx} x=12(y1/3+y−1/3)=y2/3+12y1/3x = \frac{1}{2}(y^{1/3} + y^{-1/3}) = \frac{y^{2/3} + 1}{2y^{1/3}} xy=y2/3+12y1/3y=y2/3+12y4/3\frac{x}{y} = \frac{y^{2/3} + 1}{2y^{1/3} y} = \frac{y^{2/3} + 1}{2y^{4/3}} dydx=6y4/3y2/3−1\frac{dy}{dx} = \frac{6 y^{4/3}}{y^{2/3} - 1} xy⋅dydx=y2/3+12y4/3⋅6y4/3y2/3−1=6(y2/3+1)2(y2/3−1)=3y2/3+1y2/3−1\frac{x}{y} \cdot \frac{dy}{dx} = \frac{y^{2/3} + 1}{2y^{4/3}} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} = \frac{6(y^{2/3} + 1)}{2(y^{2/3} - 1)} = 3 \frac{y^{2/3} + 1}{y^{2/3} - 1}. This matches the thought block calculation.

First term: (x2−1)y⋅d2ydx2\frac{(x^2 - 1)}{y} \cdot \frac{d^2y}{dx^2} x2−1=14(y1/3−y−1/3)2=14(y2/3−1)2y2/3x^2 - 1 = \frac{1}{4}(y^{1/3} - y^{-1/3})^2 = \frac{1}{4} \frac{(y^{2/3} - 1)^2}{y^{2/3}}. x2−1y=14y(y2/3−1)2y2/3=(y2/3−1)24y5/3\frac{x^2 - 1}{y} = \frac{1}{4y} \frac{(y^{2/3} - 1)^2}{y^{2/3}} = \frac{(y^{2/3} - 1)^2}{4y^{5/3}}. d2ydx2=24y5/3(y2/3−2)(y2/3−1)3\frac{d^2y}{dx^2} = \frac{24 y^{5/3} (y^{2/3} - 2)}{(y^{2/3} - 1)^3}. (x2−1)y⋅d2ydx2=(y2/3−1)24y5/3⋅24y5/3(y2/3−2)(y2/3−1)3=244(y2/3−1)2(y2/3−1)3y5/3y5/3(y2/3−2)\frac{(x^2 - 1)}{y} \cdot \frac{d^2y}{dx^2} = \frac{(y^{2/3} - 1)^2}{4y^{5/3}} \cdot \frac{24 y^{5/3} (y^{2/3} - 2)}{(y^{2/3} - 1)^3} = \frac{24}{4} \frac{(y^{2/3} - 1)^2}{(y^{2/3} - 1)^3} \frac{y^{5/3}}{y^{5/3}} (y^{2/3} - 2) =61y2/3−1(y2/3−2)=6y2/3−2y2/3−1= 6 \frac{1}{y^{2/3} - 1} (y^{2/3} - 2) = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1}. This also matches the thought block calculation.

E=6y2/3−2y2/3−1+3y2/3+1y2/3−1E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + 3 \frac{y^{2/3} + 1}{y^{2/3} - 1} E=6(y2/3−2)+3(y2/3+1)y2/3−1=6y2/3−12+3y2/3+3y2/3−1=9y2/3−9y2/3−1=9(y2/3−1)y2/3−1=9E = \frac{6(y^{2/3} - 2) + 3(y^{2/3} + 1)}{y^{2/3} - 1} = \frac{6y^{2/3} - 12 + 3y^{2/3} + 3}{y^{2/3} - 1} = \frac{9y^{2/3} - 9}{y^{2/3} - 1} = \frac{9(y^{2/3} - 1)}{y^{2/3} - 1} = 9.

The calculation seems correct.

Let's check for any special cases or constraints. The denominator y2/3−1y^{2/3} - 1 appears in the derivatives. This is zero if y2/3=1y^{2/3} = 1, which means y=±1y = \pm 1. If y=1y=1, 2x=11/3+1−1/3=1+1=22x = 1^{1/3} + 1^{-1/3} = 1 + 1 = 2, so x=1x=1. If y=−1y=-1, 2x=(−1)1/3+(−1)−1/32x = (-1)^{1/3} + (-1)^{-1/3}. If we consider the real cube root, this is −1+(−1)=−2-1 + (-1) = -2, so x=−1x=-1. The derivatives are undefined at (x,y)=(1,1)(x, y) = (1, 1) and (−1,−1)(-1, -1). However, the question asks for the value of the expression, which is a combination of these derivatives. The cancellation 9(y2/3−1)y2/3−1\frac{9(y^{2/3} - 1)}{y^{2/3} - 1} suggests that the result is independent of yy (and thus xx) as long as y2/3−1≠0y^{2/3} - 1 \neq 0.

The question is well-defined for values of yy where y2/3≠1y^{2/3} \neq 1. The expression simplifies to a constant value 9.

The final answer is 9\boxed{9}.

Explanation of the solution:

  1. Differentiate the given relation 2x=y1/3+y−1/32x = y^{1/3} + y^{-1/3} with respect to xx to find dydx\frac{dy}{dx} in terms of yy.
  2. Differentiate the equation for dydx\frac{dy}{dx} again with respect to xx to find d2ydx2\frac{d^2y}{dx^2} in terms of yy and dydx\frac{dy}{dx}.
  3. Express xx and x2−1x^2 - 1 in terms of yy.
  4. Substitute the expressions for xx, x2−1x^2 - 1, dydx\frac{dy}{dx}, and d2ydx2\frac{d^2y}{dx^2} in terms of yy into the given expression (x2−1)y⋅d2ydx2+xy⋅dydx\frac{(x^2 - 1)}{y} \cdot \frac{d^2y}{dx^2} + \frac{x}{y} \cdot \frac{dy}{dx}.
  5. Simplify the resulting expression in terms of yy. The terms involving yy should cancel out, leaving a constant value.

Step 1: 2=13y−2/3dydx−13y−4/3dydx  ⟹  dydx=6y4/3y2/3−12 = \frac{1}{3} y^{-2/3} \frac{dy}{dx} - \frac{1}{3} y^{-4/3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{6 y^{4/3}}{y^{2/3} - 1}. Step 2: Differentiating 6=(y−2/3−y−4/3)dydx6 = (y^{-2/3} - y^{-4/3}) \frac{dy}{dx} w.r.t. xx gives (y2/3−1)d2ydx2=23y(y2/3−2)(dydx)2(y^{2/3} - 1) \frac{d^2y}{dx^2} = \frac{2}{3y} (y^{2/3} - 2) (\frac{dy}{dx})^2. Step 3: x=12(y1/3+y−1/3)x = \frac{1}{2}(y^{1/3} + y^{-1/3}), x2−1=14(y1/3−y−1/3)2=(y2/3−1)24y2/3x^2 - 1 = \frac{1}{4}(y^{1/3} - y^{-1/3})^2 = \frac{(y^{2/3} - 1)^2}{4y^{2/3}}. Step 4: Substitute the expressions into E=(x2−1)yd2ydx2+xydydxE = \frac{(x^2 - 1)}{y} \frac{d^2y}{dx^2} + \frac{x}{y} \frac{dy}{dx}. First term: (x2−1)yd2ydx2=(y2/3−1)24y5/3⋅24y5/3(y2/3−2)(y2/3−1)3=6y2/3−2y2/3−1\frac{(x^2 - 1)}{y} \frac{d^2y}{dx^2} = \frac{(y^{2/3} - 1)^2}{4y^{5/3}} \cdot \frac{24 y^{5/3} (y^{2/3} - 2)}{(y^{2/3} - 1)^3} = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1}. Second term: xydydx=12(y1/3+y−1/3)y⋅6y4/3y2/3−1=(y1/3+y−1/3)3y4/3y(y2/3−1)=(y2/3+1)2y4/36y4/3y2/3−1=3y2/3+1y2/3−1\frac{x}{y} \frac{dy}{dx} = \frac{\frac{1}{2}(y^{1/3} + y^{-1/3})}{y} \cdot \frac{6 y^{4/3}}{y^{2/3} - 1} = \frac{(y^{1/3} + y^{-1/3}) 3 y^{4/3}}{y(y^{2/3} - 1)} = \frac{(y^{2/3} + 1)}{2y^{4/3}} \frac{6 y^{4/3}}{y^{2/3} - 1} = 3 \frac{y^{2/3} + 1}{y^{2/3} - 1}. Step 5: E=6y2/3−2y2/3−1+3y2/3+1y2/3−1=6(y2/3−2)+3(y2/3+1)y2/3−1=6y2/3−12+3y2/3+3y2/3−1=9y2/3−9y2/3−1=9(y2/3−1)y2/3−1=9E = 6 \frac{y^{2/3} - 2}{y^{2/3} - 1} + 3 \frac{y^{2/3} + 1}{y^{2/3} - 1} = \frac{6(y^{2/3} - 2) + 3(y^{2/3} + 1)}{y^{2/3} - 1} = \frac{6y^{2/3} - 12 + 3y^{2/3} + 3}{y^{2/3} - 1} = \frac{9y^{2/3} - 9}{y^{2/3} - 1} = \frac{9(y^{2/3} - 1)}{y^{2/3} - 1} = 9.

The final answer is 9.

The final answer is 9\boxed{9}.