Question

The function y1(x) = x is a solution of x^2y''+xy'-y=0. Find the general solution of the differential equation x^2y''+xy'-y=2x.

Answer 1

When I plugged in the equations into WA, I did not get that as a solution, so I am really confused on what to do.

Answer 2

http://www.wolframalpha.com/input/?i=x%5E2y%27%27%2Bxy%27-y%3D2x
Interesting ... there seem to be complex term on the solution

Answer 3

the constants of integration could be complex valued, just replace it by c_2, rest should be fine.

Answer 4

there is an equation that this problem references but I do not understand how it applies.

Answer 5

this is Euler Cauchy equation ... there is along article on wikipedia about how to solve it ... also since one solution is given, try using the method of reduction of order to solve it
http://en.wikipedia.org/wiki/Reduction_of_order

Answer 6

that example on there is kind of confusing.

Answer 7

Let, \( y_2 = v(x) x \) be the other solution, not put that solution on your DE

Answer 8

x^ (v(x) x) '' + ... = 2x

Answer 9

is v(x) the homogeneous equation

Answer 10

or i guess i dont understand what v(x) actually is

Answer 11

no ... it's just assumption that your other solution is of that form.

Answer 12

so how do you solve for the other solution when you dont even know what v(x) actually is?

Answer 13

hold on for a while.

Answer 14

It the problem says to reference another problem which states.

Let y=y1(x) be a solution of (H):y''+p(x)y'+q(x)y=0 where p and q are continuous function on an interval I. Let a be in I and assume that y1(x) not equal to 0 on I. Set
y2(x) = y1(x)integral (e^-(integral p(u)du))/y1^2(t)) dt.
Show that y2 is a solution of (H) and that y1 and y2 are linearly independent.

I see that it is the same kind of problem, but I dont see how this helps in anyway.

Answer 15

you should be getting in this form
http://upload.wikimedia.org/math/2/7/e/27ec56e8d0dc1f64f12feb692f4e5660.png
which is a linear DE

Answer 16

Of course substituting \( y_2 = v x \) gives
\[x^2 (v'' x + 2 v' )+x(v'x+v) - vx = 2x \\
x^3 v'' +3x^2 v' = 2x \\
x^2 v'' + 3x v' = 2\]
Let v' = u, we get \( x^2 u' + 3x u = 2\) which is linear ... now solve for 'u' then sovle for 'v' .. and put that thing in \(y = vx \) and get your final solution.

Answer 17

am i just going to get u by itself and then integrate?

Answer 18

yep ... we know v' = u, so we can find v from u.

Answer 19

this is pretty hectic ... you should have directly done this which is much simpler
http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

Answer 20

so i integrate 1/3(2-x^2)x^-1 and get

x^2/3 - x^4/12

Answer 21

but im not sure where the u went?

Answer 22

if u = (2-x^2(u'))/3x i guess i ignored the u, but you cant do that