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The auxiliary equation for the given differential equation has complex roots. Find a general solution. y''-10y' 29y=0

User Sean Quinn
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Answer:


y = Acos5x - Bsin5x

Explanation:

Given the differential equation y''-10y'+29y=0

First, we need to rewrite it as an auxiliary equation as shown:

Let y'' = m²y and y' = my

Substitute the values into the general equation

m²y-10my+29y = 0

Factor out y:

(m²-10m+29)y = 0 [The auxiliary equation]

Solve the auxiliary equation and find the roots of the equation

m²-10m+29 = 0

m = -b±√(b²-4ac)/2a

a = 1, b = -10, c = 29

m = -10±√(10²-4(1)(29))/2(1)

m = -10±√(100-116)/2

m = -10±√-16/2

m = (-10±4i)/2

m = -10/2 + 4i/2

m = -5+2i

Comparing the complex number with a+bi, a = -5 and b = 2

The general solution for complex solution is expressed as:


y = Acosax + Bsinax

Substitute the value of a in the equation


y = Acos(-5)x + Bsin(-5)x\\y = Acos5x-Bsin5x

Hence the general solution to the differential equation is
y = Acos5x - Bsin5x

User Haydee
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