The function y(x) represented by x=sin t,y=a | vedclass

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(B) Given $x = \sin t$,we have $\frac{dx}{dt} = \cos t$. Given $y = a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}$. Then $\frac{dy}{dt} = a \sqrt{2} e^{t \sqrt

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$2$

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(B) Given $x = \sin t$,we have $\frac{dx}{dt} = \cos t$. Given $y = a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}$. Then $\frac{dy}{dt} = a \sqrt{2} e^{t \sqrt{2}} - b \sqrt{2} e^{-t \sqrt{2}} = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$. Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})}{\cos t}$. Let $y' = \frac{dy}{dx}$. Then $y' \cos t = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$. Differentiating both sides with respect to $t$: $y'' \cos t - y' \sin t = \sqrt{2} \cdot \sqrt{2}(a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}) = 2y$. Since $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$,we have $y'' \sqrt{1 - x^2} - y' \frac{dx}{dt} \cdot \frac{1}{\cos t} \cdot \sin t = 2y$. Actually,using $y' = \frac{dy}{dx}$,we have $\frac{d}{dx}(y' \cos t) = \frac{d}{dx}(\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}}))$. $y'' \cos t + y'(-\sin t) \frac{dt}{dx} = 2y \frac{dt}{dx}$. Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we get $y'' \cos t - y' \frac{\sin t}{\cos t} = 2y \frac{1}{\cos t}$. Multiplying by $\cos t$: $y'' \cos^2 t - y' \sin t = 2y$. Substituting $\cos^2 t = 1 - x^2$ and $\sin t = x$: $(1 - x^2) y'' - x y' = 2y$. Comparing with $(1 - x^2) y'' - x y' = ky$,we get $k = 2$.

The function $y(x)$ represented by $x=\sin t$,$y=a e^{t \sqrt{2}}+b e^{-t \sqrt{2}}$,$t \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ satisfies the equation $(1-x^2) y^{\prime \prime}-x y^{\prime}=k y$,then the value of $k$ is

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