The general solution to the second-order differential equation y′′+10y=0 is in the form y(x)=c1cosβx+c2sinβx. Find the value of β, where β>0.

we will differentiate the y(x) equation twice to get a second order differential equation.
We compare our second order differential equation with the Second order differential equation specified in the problem to get the value of β

y(x)=c1cosβx+c2sinβx

we use the derivative of a sum rule to differentiate since we have an addition sign in our equation.

Also when differentiating Cosβx and Sinβx we should note that this involves function of a function. so we will differentiate βx in each case and multiply with the differential of c1cosx and c2sinx respectively.

lastly the differential of sinx= cosx and for cosx = -sinx.

Knowing all these we can proceed to solving the problem.

y=c1cosβx+c2sinβx

y'= β×c1×-sinβx+β×c2×cosβx

y'=-c1βsinβx+c2βcosβx

y''=β×-c1β×cosβx + (β×c2β×-sinβx)

y''= -c1β²cosβx -c2β²sinβx

factorize -β²

y''= -β²(c1cosβx +c2sinβx)

y(x)=c1cosβx+c2sinβx

therefore y'' = -β²y

y''+β²y=0

now we compare this with the second order D.E provided in the question

y''+10y=0

this means that β²y=10y

β²=10

B=√10 or 3.16(2 d.p)