The auxiliary equation for the given differential equation has complex roots. Find a general solution. y''-4y' +7y =0

y = Ae^2xcos√3 x + Be^2xsin√3 x

Step-by-step explanation:

Given the differential equation

y''-4y' +7y =0

Let my = y' and m²y =y''

The auxiliary equation is expressed as;

m²y - 4my +7y = 0

(m²-4m+7)y =0

Divide through by y;

m²-4m+7= 0

Factorize the auxiliary equation

m = 4±√16-4(7)/2

m = 4±√-12/2

m = 4±2√3 i/2

m = 2+√3i

From the complex number a = 2(real part)

b = √3 (imaginary part)

Substitute into the general solution

y = Ae^axcosbx + Be^axsinbx

y = Ae^2xcos√3 x + Be^2xsin√3 x

days of the week

step-by-step explanation:

javier earned more money then kyle, so that won't always be a constant term, but the days of the week do not change so that term is always constant.

edit: i meant

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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