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!
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
To solve this problem, we can start by assuming a solution of the form y = x^r, where r is a constant to be determined. We do this because the hint suggests trying this form.
To find the first and second derivatives of y with respect to x, we need to apply the power rule from calculus. The first derivative is given by:
y' = r*x^(r-1),
and the second derivative is given by:
y'' = r(r-1)*x^(r-2).
Now, substitute y, y', and y'' into the original differential equation:
x^(2)y'' - xy' = 0.
Substituting the derivatives into the equation, we have:
x^(2)(r(r-1)*x^(r-2)) - x(x^(r-1)) = 0.
Simplifying this equation, we get:
r(r-1)*x^r - x^r = 0.
Now, factor out the common x^r term:
x^r(r(r-1) - 1) = 0.
Here, we have two solutions:
x^r = 0, which corresponds to r = 0,
(r(r-1) - 1) = 0, which simplifies to r^2 - r - 1 = 0.
The first solution, r = 0, gives us y = x^0 = 1.
Now, for the second solution, we solve the quadratic equation r^2 - r - 1 = 0. Using the quadratic formula, we find:
r = [1 ± sqrt(1 + 4*1*1)] / 2.
r = [1 ± sqrt(5)] / 2.
Hence, the two solutions for r are:
r1 = (1 + sqrt(5))/2,
r2 = (1 - sqrt(5))/2.
Thus, we have found two solutions: y1 = 1 and y2 = x^[(1 + sqrt(5))/2].
To show that these solutions are linearly independent, we need to show that no multiple of one solution can be written as a constant multiple of the other. In other words, there is no constant C such that y2 = C*y1 for all x.
To do this, we can compute the Wronskian of these two functions, given by:
W(y1, y2) = y1*y2' - y1'*y2.
Calculating the derivatives and substituting into the Wronskian formula, we get:
W(y1, y2) = x^[(1 + sqrt(5))/2] - [(1 + sqrt(5))/2]x^[(1 + sqrt(5))/2 - 1] = x^[(1 + sqrt(5))/2] - x^[(1 - sqrt(5))/2].
Since the exponents in the Wronskian are different, there is no way to factor out an x term from both terms, hence W(y1, y2) is nonzero for all x.
Therefore, y1 = 1 and y2 = x^[(1 + sqrt(5))/2] are linearly independent.
The general solution of the differential equation is given by:
y(x) = C1*y1 + C2*y2,
where C1 and C2 are constants, and y1 = 1 and y2 = x^[(1 + sqrt(5))/2] are the solutions we found.