Question

y=A cos nx + B sin nx prove that d^2y/dx^2 + n^2y=0

Answer 1

$\textbf{Answer :}$

Given,

$ y = A cos nx + B sin nx $

Differentiating both sides with respect to x, we get

dy/dx = - nA sin nx + nB cos nx

Again, Differentiating with respect to x, we get

$\frac{ {d}^{2}y }{d {x}^{2} } = - {n}^{2} A \: cos \: nx - {n}^{2} B \: sin \: nx \\ \\ Or, \: \: \frac{ {d}^{2} y}{d {x}^{2} } + {n}^{2} (A \: cos \: nx \: + B \: cos \: nx) = 0 \\ \\ Or, \: \: \frac{ {d}^{2}y }{d {x}^{2} } + {n}^{2} y = 0 \\ \\ Hence, \: \: proved.$

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