hcortez5547 hcortez5547
  • 25-11-2020
  • Mathematics
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The auxiliary equation for the given differential equation has complex roots. Find a general solution.

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

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abidemiokin
abidemiokin abidemiokin
  • 26-11-2020

Answer:

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

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