(1 point) Find the general solution to the homogeneous differential equation. 2y2−20y+136y=0 Use c1 and c2 in your answer to denote arbitrary constants, and enter them as c1 and c2. y()= ?

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Your differential equation is not displayed well. It though looks like this:

2d²y/dx² - 20dy/dx + 136y = 0

If this is not the differential equation, the method of solving this would still be used in solving the correct one.

We first write an auxiliary equation to the differential equation.

The auxiliary equation is:

2m² - 20m + 136 = 0

Dividing by 2, we have

m² - 10m + 68 = 0

Next, we solve the auxiliary equation to obtain the values of m.

Solving using the quadratic formula

m = [-b ± √(b² - 4ac)]/2a

Where a = 1, b = -10, and c = 68

m = [10 ± √(100 - 272)]/2

= 5 ± (1/2)√(-172)

= 5 ± (1/2)i√172

= 5 ± 6.6i

For solutions of the form a ± ib, the complimentary solution is

y = e^(ax)[C1cosbx + C2sinbx]

Therefore, the complimentary solution is

y = e^(5x)[C1cos(6.6x) + C2sin(6.6x)]

there are no short passages to look at