Answer to Question #317847 in Calculus for Segun

Question

Given that y= acoskx + bsinkx, show that d²y/dx² + k²y =0

Answer 1

$y(x)=acos(kx)+bsin(kx)$

$y'(x)=-a\times ksinkx+b\times kcos(kx)$

$y"(x)=-a\times k^2\times cos(kx)-b\times k^2sin(kx)$

$=-k^2(acos(kx)+bsin(kx))$

From this its clear that,

$y"(x)=-k^2\times y(x)$

y"+k²y=0

Hence,

$\frac{d^2y}{dx^2}+k^2y=0$