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

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

answer: i literally just answered this question but they want me to do it again

step-by-step explanation: