. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd a second solution y2(x) of the homogeneous equation. x 2 y 00 7xy0 + 16y = 0; y1 = x 4

The question is:

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

x²y'' - 7xy' + 16y = 0; y1 = x^4

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

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

Let y2 = uy1

=> y2 = ux^4

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'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

So

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

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step-by-step explanation:

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