please solve: x^2•y"-7xy'+16y=(lux+1)•x^4

Question

x^2•y"-7xy'+16y=(lux+1)•x^4

Answer 1

Let y=x^k then k^2-8k+16_≈(k-4)^2=0

Answer 2

→k=4(double degeneracy).

Answer 3

1st solution of the homogenous: y,=x^4

Answer 4

The 2nd is obtained by lagrange's reduction of order:

Answer 5

y2=x^4.lux.

Answer 6

That is Yh=C1Y1+C2Y2

Answer 7

To find a particular solution satisfying the given we evaluate the wrouskian determinants:

Answer 8

|w|=Y1•Y2-Y1’•Y2=x^7, |w1|=-Y2 , |w2|=Y1

Answer 9

Let r(x)=1/x^2.(lux+1•x^4

Answer 10

→r(x)=x^2•(lux+1)

Answer 11

Now let E1(x)_=|w1|/|w|•r(x) & E2(x)_=|w2|/|w|•r(x)

Answer 12

If Yp is the particular solution then the Y satisfying the given non- homogenous EULER-CAUCHY equation is Y=Yh+Yp

Answer 13

Yp=Y1•{E1dx+Y2}E2dx→Yp=-x^4•(1/3lu^3x+1/2lu^2x)+x^4•(1/2lu^3x+lu^2x)

Answer 14

→Yp=1/6(3+lux)•x^4•lu^2x

Answer 15

Therefore thw Y is:

Answer 16

Y=[1/6lu^3x+1/2lu^2x+C2lux+C1]•x^4