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B
y′′=2z
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C
z′′=−2y
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D
z′=−y
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Solution
The correct options are Ay′′=2z Cz′′=−2y y=exsinxy′=exsinx+excosxy′′=exsinx+excosx+excosx−exsinx=2excosx=2zz=excosxz′=excosx−exsinxz′′=excosx−exsinx−exsinx−excosx=−2exsinx=−2y