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,

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