If $y^2=a^2\cos^2x+b^2\sin^2x$, show that $y+\dfrac{d^2y}{dx^2}=\dfrac{a^2b^2}{y^3}$

If y²=a²cos²x+b²sin²x, show that y+(d²y)/(dx²)=(a²b²)/(y³)

MISCELLANEOUS EXERCISE 1 (II) · Q247

y^2=a^2\cos^2x+b^2\sin^2x

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Differentiate $y^2=a^2\cos^2x+b^2\sin^2x$ w.r.t. $x$ :

2yy'=-2a^2\cos x\sin x+2b^2\sin x\cos x=2\sin x\cos x(b^2-a^2)=(b^2-a^2)\sin2x.

So $yy'=\dfrac{(b^2-a^2)}{2}\sin2x$ . Differentiate again:

y'^2+yy''=(b^2-a^2)\cos2x.

Also, from $y^2=a^2\cos^2x+b^2\sin^2x=a^2+(b^2-a^2)\sin^2x$ , we get $(b^2-a^2)\sin^2x=y^2-a^2$ , and similarly $(b^2-a^2)\cos^2x=b^2-y^2$ (since $y^2=b^2-(b^2-a^2)\cos^2x$ ). So

(b^2-a^2)\cos2x=(b^2-a^2)(\cos^2x-\sin^2x)=(b^2-y^2)-(y^2-a^2)=a^2+b^2-2y^2.

Also $y'^2=\dfrac{(yy')^2}{y^2}=\dfrac{(b^2-a^2)^2\sin^2x\cos^2x}{y^2}=\dfrac{(y^2-a^2)(b^2-y^2)}{y^2}$ (using the two boxed relations above).

$yy''=a^2+b^2-2y^2-y'^2=a^2+b^2-2y^2-\frac{(y^2-a^2)(b^2-y^2)}{y^2}.$ …

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Q.If $y^2=a^2\cos^2x+b^2\sin^2x$ , show that $y+\dfrac{d^2y}{dx^2}=\dfrac{a^2b^2}{y^3}$

Concept understanding — Second Order Derivative

Q.If y2=a2cos⁡2x+b2sin⁡2xy^2=a^2\cos^2x+b^2\sin^2xy2=a2cos2x+b2sin2x, show that y+d2ydx2=a2b2y3y+\dfrac{d^2y}{dx^2}=\dfrac{a^2b^2}{y^3}y+dx2d2y​=y3a2b2​

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Q.If $y^2=a^2\cos^2x+b^2\sin^2x$ , show that $y+\dfrac{d^2y}{dx^2}=\dfrac{a^2b^2}{y^3}$

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