Question

consider the equation x^2 y''-7xy'+15y=0, show that y=x^3(x>0) is a solution of the equation and find a second independent solution.

Answer 1

hint :
this is Euler's homogeneous equation
he auxillary equation is
D'(D'-1) -7D' +15=0

hence
(D'-3)(D'-5)=0

so check if the other solution is of the form y=kx^5 , k being constant ...

Answer 2

x^2 y''-7xy'+15y=0
Using the power rule calculate the 1st and 2nd derivatives:
\[y' = 3 x ^{2} \]
\[y'' = 6x\]

So, substitute for y, y' and y'':
\[x ^{2}.(6x) - 7x(3x ^{2}) + 15(x ^{3}) = 0\]
\[6x ^{3} - 21x ^{3} + 15x ^{3} = 0\]
That's easy math right there.

I will defer to matricked for the other solution.