Given that y(t)=c1e6t+c2e−6ty(t)=c1e6t+c2e−6t is a solution to the differential equation y′′−36y=0y′′−36y=0, where c1c1 and c2c2 are arbitrary constants, find a function y(t)y(t) that satisfies the conditions y′′−36y=0,y′′−36y=0, y(0)=9,y(0)=9, limt→+∞y(t)=0.limt→+∞y(t)=0.

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!