Question

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

Answer 1

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.