Question

Given the diff eq:

y'' - (2/x)y' - (4/x^2)y = 0

I found two values of r such that y = x^r. These values were 4 and -1.

The general solution to this is

y = C(1)x^4 + C(2)1/x

Now I need to find the solution of the equation satisfying the initial conditions
y(1) = 2, y'(1) = -1

How do I do that?

Answer 1

I haven't checked the solution, but I'll take your word for it that it's correct.

Given that
\[y = C_1x^4+C_2x^{-1},\]
and that y(1) = 2, you have
\[2 = C_1(1)^4+C_2(1)^{-1} \;\;\;\;\;\;\;\;\;(A)\]
Also, you have
\[y' = 4C_1x^3 - C_2x^{-2},\]
and y'(1)=-1, so
\[1 = 4C_1(1)^3 - C_2(1)^{-2}. \;\;\;\;\;(B)\]
How do you solve for C_1 and C_2, using equations (A) and (B)?

Answer 2

so c1 is 1/5 and c2 is 9/5?

Answer 3

for equation (b) i assumed you meant to put -1 = ... right.

Answer 4

Yes, my mistake. Thanks for pointing it out.

Answer 5

so are my c1 and c2 correct?