The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution of the given nonhomogeneous equation. y'' − 9y = 4; y1 = e−3x

The second solution y2 is

4/3 + (1/6)C1e^(3x) + C2e^(3x)

Step-by-step explanation:

Given the nonhomogeneous differential equation

y'' - 9y = 4

And a solution: y1 = e^(-3x)

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ue^(-3x)

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. e^(-3x) + u. -3e^(-3x)

y2'' = u''. e^(-3x) + u'. -3e^(-3x) + u'. -3e^(-3x) + u. 9e^(-3x)

= u''. e^(-3x) - u'. 6e^(-3x) + u. 9e^(-3x)

Now, using these values in the original equation,

[u''. e^(-3x) - u'. 6e^(-3x) + u. 9e^(-3x)] - 9[ue^(-3x)] = 4

[u'' - 6u' + (9 - 9)u]e^(-3x) = 4

u'' - 6u' = 4e^(3x)

Let w = u'

Then w' = u''

So

w' - 6w = 4e^(3x)

Let the integrating factor be

e^(integral of -6)dx

= e^(-6x)

Multiply both sides of the equation by the integrating factor.

e^(-6x)(w' - 6w) = 4e^(3x).e^(-6x)

d[we^(-6x)] = 4e^(-3x)

Integrating both sides

we^(-6x) = 4e^(-3x) + C1

w = 4e^(3x) + C1e^(6x)

But w = u'

So,

u' = 4e^(3x) + C1e^(6x)

Integrating this

u = (4/3)e^(3x) + (1/6)C1e^(6x) + C2

Since

y2 = uy1

We have

y2 = [(4/3)e^(3x) + (1/6)C1e^(6x) + C2e^(6x)]. e^(-3x)

= 4/3 + (1/6)C1e^(3x) + C2e^(3x)

Therefore, the second solution is

y2 = 4/3 + (1/6)C1e^(3x) + C2e^(3x)

the answer

b. they are parallel and will never intersect.