Questions a, b, and c below for the following info: I have attempted to answer,
ID: 2982315 • Letter: Q
Question
Questions a, b, and c below for the following info:
I have attempted to answer, but it really does not seem to be correct?? A detailed answer with explanation would be greatly appreciated!
Thanks!
____________________________________________________________
y1=x^2
y2=x^2lnx
y''-(3/x)y'+(4/x^2)y=4
a)Show that y1 and y2 form a fundamental set of solutions of the reduced equation
b)Find a particular solution of the given equation
c)Find a general solution of the given equation
_____________________________________________________________
a)
W =(matrix) x^2 x^2lnx
2x 2xlnx+x
W= 2x^3 lnx +x^3 - 2x^3 lnx
= x^3 ~= 0
b)
y = Ax^2 + Bx + C
y' = 2Ax + B
y'' = 2A
plug into DE
x^2(2A-3A+4A) +x(3B+4B)+4C=4
3A=0 7B=0 4C=4
A=0 B=0 C=1
z=1
c)
x^2(y''-(3/x)y'+(4/x^2)y=o)
x^2y''-3xy'+4y=0
x^2r^2 - 3xr + 4r = 0
(xr + 1)(xr - 4)
r = -1/x, 4/x
y = C1 e^(-1/x) + C2 e^(4/x) + 1
Explanation / Answer
i believe that a and c are wrong i solved them below.
____________________________________________________________
y1=x^2
y2=x^2lnx
y''-(3/x)y'+(4/x^2)y=4
a)Show that y1 and y2 form a fundamental set of solutions of the reduced equation
b)Find a particular solution of the given equation
c)Find a general solution of the given equation
_____________________________________________________________
a)
y''-(3/x)y'+(4/x^2)y=4
y1 = x^2
y1' = 2x
y1'' = 2
put in above DE
2 - (3/x) 2x + (4/x^2) x^2 = 0
2 -3*2 + 4 = 0 so Y1 is solution
x^2y''-3xy'+4y=0
y2 = x^2 lnx
y2' = x^2/x + 2x lnx = x + 2x lnx
y2'' = 1 + 2x/x + 2lnx = 3 + 2lnx
x^2 ( 3 + 2 lnx ) - 3x ( x + 2x lnx ) + 4 x^2 lnx =
= 3x^2 + 2x^2 lnx - 3x^2 -6x^2 lnx + 4x^2 lnx = 0
so y2 also forms fundamental solution.
b)
y = Ax^2 + Bx + C
y' = 2Ax + B
y'' = 2A
plug into DE
x^2(2A-3A+4A) +x(3B+4B)+4C=4
3A=0 7B=0 4C=4
A=0 B=0 C=1
z=1
c)
x^2(y''-(3/x)y'+(4/x^2)y=o)
x^2y''-3xy'+4y=0
this is cauchy euler DE
with b = -3 and c = 4
thus
auxliry equation is
r^2 + (-3-1) r + 4 = 0
r^2 -4r+4 = 0
(r-2)^2 = 0
r = 2,2 now
solution will be
y = ( C1 + C2 ln(x) ) x^2
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