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If \mathrm{2 f(\sin x)+f(\cos x)=x} for all \mathrm{x \in \mathbf{R}} then \mathrm{y=f(x)} satisfies

Option: 1

\mathrm{\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=0}


Option: 2

\mathrm{x^{2} y^{\prime \prime}-\left(1-x^{2}\right) y^{\prime}=0}


Option: 3

\mathrm{\left(1+x^{2}\right) y^{\prime \prime}+x y^{\prime}=0}


Option: 4

\mathrm{\left(1-x^{2}\right) y^{\prime \prime}+x y^{\prime}=0}


Answers (1)

best_answer

Replacing \mathrm{x} by \mathrm{\pi / 2-x} in the given equation; we get

\mathrm{2 f(\cos x)+f(\sin x)=\pi / 2-x}

Solving this equation with the given equation, we obtain

\mathrm{f(\cos x)=\pi / 3-x \Rightarrow y=f(x)=\pi / 3-\cos ^{-1} x}
\mathrm{\Rightarrow \quad y^{\prime}=-\frac{1}{\sqrt{1-x^{2}}} \quad \Rightarrow \quad y^{\prime 2}\left(1-x^{2}\right)=1}

Differentiating again we have

\begin{aligned} & 2 y^{\prime} y^{\prime \prime}\left(1-x^{2}\right)-2 x y^{\prime 2}=0 \\ & \Rightarrow \quad\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=0 . \end{aligned}

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