If y=a sin x+b cos x,then y^2+left(frac{d y}{ | vedclass

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(C) Given,$y=a \sin x+b \cos x$ Differentiating with respect to $x$,we get: $\frac{d y}{d x} = a \cos x - b \sin x$ Now,consider the expression $y^2 +

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(C) Given,$y=a \sin x+b \cos x$ Differentiating with respect to $x$,we get: $\frac{d y}{d x} = a \cos x - b \sin x$ Now,consider the expression $y^2 + \left(\frac{d y}{d x}\right)^2$: $y^2 + \left(\frac{d y}{d x}\right)^2 = (a \sin x + b \cos x)^2 + (a \cos x - b \sin x)^2$ Expanding the squares: $= (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$ Grouping the terms: $= a^2(\sin^2 x + \cos^2 x) + b^2(\sin^2 x + \cos^2 x)$ Since $\sin^2 x + \cos^2 x = 1$: $= a^2(1) + b^2(1) = a^2 + b^2$ Since $a$ and $b$ are constants,$a^2 + b^2$ is a constant. Thus,the expression is a constant.

If $y=a \sin x+b \cos x$,then $y^2+\left(\frac{d y}{d x}\right)^2$ is a

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