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If Y = 2 Sin X + 3 Cos X, Show that D 2 Y D X 2 + Y = 0 ? - Mathematics

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Question

If y = 2 sin x + 3 cos x, show that \[\frac{d^2 y}{d x^2} + y = 0\] ?

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Solution

Here,

\[y = 2 \sin x + 3 \cos x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = 2\cos x - 3 \sin x\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = - 2 \ sin \ x - 3 \ cos \ x \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - \left( 2 \sin x + 3 \ cos x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - y\]
\[ \Rightarrow \frac{d^2 y}{d x^2} + y = 0\]

Hence proved.

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Chapter 12: Higher Order Derivatives - Exercise 12.1 [Page 16]

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RD Sharma Mathematics [English] Class 12
Chapter 12 Higher Order Derivatives
Exercise 12.1 | Q 6 | Page 16

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