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If $y=2 \sin x+3 \cos x$ and $y+A \frac{d^2 y}{d x^2}=B$, then the values of $A$, $\mathrm{B}$ are respectively
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The correct answer is:
1,0
$$
\begin{aligned}
& y=2 \sin x+3 \cos x \\
& \therefore \frac{d y}{d x}=2 \cos x-3 \sin x \\
& \therefore \frac{d^2 y}{d x^2}=-2 \sin x-3 \cos x=-(2 \sin x+3 \cos x)=-y \\
& \therefore y+\frac{d^2 y}{d x^2}=0
\end{aligned}
$$
We have $y+A \frac{d^2 y}{d x^2}=B \Rightarrow A=1, B=0$
\begin{aligned}
& y=2 \sin x+3 \cos x \\
& \therefore \frac{d y}{d x}=2 \cos x-3 \sin x \\
& \therefore \frac{d^2 y}{d x^2}=-2 \sin x-3 \cos x=-(2 \sin x+3 \cos x)=-y \\
& \therefore y+\frac{d^2 y}{d x^2}=0
\end{aligned}
$$
We have $y+A \frac{d^2 y}{d x^2}=B \Rightarrow A=1, B=0$
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