# Unclear concept on simple 2nd order differential eq

1. Nov 10, 2015

### davon806

1. The problem statement, all variables and given/known data
Hi,I am learning to solve 2nd-order differential eq.
Suppose I have a equation
dy/dx - 3x = 0...........(1)
Then dy/dx = 3x -----> x = 3(x^2)/2
Now if I have a 2nd order ODE such that:
d^2y/dx^2 = 3.............(2)
Then it could be solved by integrating both sides wrt x twice,which yields
y = 3(x^2)/2 + Ax + B

Now,consider the case:
d^2y/dx^2 -6dy/dx + 9 = 0.........(3)
I know it could be solved by using the idea of auxiliary equation(Putting y = Ae^(sx) into the initial eq)

But,why it's not possible to solve (3) by using direct integration on both sides as in (1) and (2)?
I can illustrate it here:
d^2y/dx^2 = 6dy/dx - 9,then I integrate both sides wrt x
dy/dx = 6 -9x + A
y = 6x -9(x^2)/2 + Ax + B
It's obviously incorrect,the (6+A)x term vanishes in the 2nd order derivative.Can someone tell me what's wrong?(I mean on the idea)

Thx
2. Relevant equations

3. The attempt at a solution
I have illustrated it in the problem statement.

2. Nov 10, 2015

### Samy_A

Didn't you lose a $y$ here?
As in: $y'=6y-9x+A$.

3. Nov 10, 2015

### davon806

Thx so much^^,a silly mistake

4. Nov 10, 2015

### Staff: Mentor

As already noted by @samy A, you lost a factor of y in the second line of your work.

Starting with y'' = 6y' - 9, if we integrate (w. respect to x), we get
y' = 6y - 9x + A

If you try the same trick again, you will have y on the left side, but will have $\int y dx$ on the right side. Since you don't know what y is in terms of x, the integral can't be calculated.