Hmmm. I'd let v = y'
Then we have
x^2 v' = xv
v' = v/x
dv/v = dx/x
log v = log x + c
v = e^c x
or,
v = c1 * x
That means y' = c1 * x
y = c1 x^2 + c2
check:
y' = 2c1 x
y" = 2c1
x^2 y" - xy' = x^2(2c1) - x(2c1 x) = 0
I'm having trouble solving this problem:
For the equation x^(2)y′′ - xy′ = 0 , find two solutions, show that they are linearly independent and find the general solution. Hint: Try y = x^(r).
I divided the equation by x^(2) first to simplify it, and then I found the first and second derivatives of y and tried plugging them back in. But I don't know where to go from there. Please help! Thank you!
1 answer
1 answer