Question

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

Answer 1

Answer:

$y = Acos5x - Bsin5x$

Step-by-step 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 $y = Acos5x - Bsin5x$