consider the equation x^2 y''-7xy'+15y=0, show that y=x^3(x0) is a solution of the equation and find a second independent solution.

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.

anonymous · Answer

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   ...

mandre · Answer

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.