Question

Find the solution of 2x^2*y"+xy'-3y=0, y(1)=1, y'(1)=4.

Answer 1

Is that Euler's formula apply?

Answer 2

I don't know, do I need to use the method that I used last time?

Answer 3

I'll be back, need eating something, give me time for eating,

Answer 4

Okay.

Answer 5

read my note, part Euler's Equation, and try it first

Answer 6

\(y=x^r\) checks out.

Answer 7

So do you use that substitution again like last time to find the general solution?

Answer 8

Yes

Answer 9

Wait a minute.

Answer 10

So I got r=-1, 3/2. Now what?

Answer 11

Now your general solution is
\[y=C_1x^{3/2}+C_2x^{-1}\]
Use the initial conditions to solve for the constants.

Answer 12

So when I plug in y(1)=1, I got c1+c2=1.

Answer 13

Right, and what about y'(1)?

Answer 14

4=c1*3/2*x^(1/2)-c2*x^-2

Answer 15

Replace x with 1...

Anyway, you have the system
\[\begin{cases}C_1+C_2=1\\\frac{3}{2}C_1-C_2=4\end{cases}\]
Solve for the constants, replace them in the general solution, and you're done!

Answer 16

Thank you, Siths, you made everything so easy.

Answer 17

@SithsAndGiggles we don't care about the coefficient of x? I mean 2 in this case.

Answer 18

You're welcome.

@Loser66, what do you mean? Did you find \(C_1=2\)?

Answer 19

nope, I talk about your first comment "y = x^r check out"

Answer 20

Yes, that's the guess. At this point, you could include the arbitrary \(C\), but the algebra is cleaner without it.

In the end, you have to remember that \(x^r\) is a particular solution and \(Cx^r\) is a general solution. Using the first is slightly more convenient.

Answer 21

on the previous post the coefficient of x is 1, you suggest let (x+c)^r because it form from the form (x+1) whose coefficient of x is 1, this case, it's 2

Answer 22

Ah, I see what you mean. The coefficients in the ODE don't change the answer.

Answer 23

Got you. Thanks a lot. I can learn from other's post, hihihi... That's great!!