The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to ?nd a second solution y2(x). y``+2y`+y=0

Step-by-step explanation:

We will use the reduction of order to solve this equation. At first, we need a solution of the homogeneus solution.

Consider the equation We will assume that the solution is of the form . If we plug this in the equation, we get

Since the exponential function is a positive function, and A should be different to zero to have non trivial solutions, we get

By using the quadratic formula, we get the solutions

So one solution of the homogeneus equation is of the form . To use the reduction of order assume that

where . We calculate the derivatives and plug it in the equation

If we rearrange the equation we get

Since is a solution of the homogeneus equation we get

If we take w = v', then w' = v''. So, in this case the equation becomes

Note that so

. Since cannot be zero, this implies

w' =0. Then, w = K (a constant). Then v' = K. So v=Kx+D where D is a constant.

So, we get that the general solution is

where C, F are constants.

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube