Question

ODE second order

d^2 y/dx^2 = y'' = xy

find the general solution,
please help, i dont know how to do..
i have already forgotten the method..
need solution

Answer 1

I wonder if it is separable? \[

\begin{array}{rcl}
\frac{d^2y}{dx^2} &=& xy \\
\frac{dy}{dx}\frac{dy}{dx} &=& xy \\
\frac{1}{y}\frac{dy}{dx}\frac{dy}{dx} &=& x \\
\int \int \frac{1}{y}\frac{dy}{dx}\frac{dy}{dx}dxdx &=& \int \int x dx dx\\


\end{array}
\]

Answer 2

Since \[
dy = \frac{dy}{dx}dx
\]I'm thinking \[
\frac{dy}{dx}\frac{dy}{dx}dxdx = \frac{dy}{dx}dydx = \frac{dy}{dx}dxdy = dydy
\]

Answer 3

If so then don't we just have: \[
\int \int \frac{1}{y}dydy = \int \int xdxdx
\]

Answer 4

@ikamashi Can you do these integrals?

Answer 5

What you have is:
y'' - xy = 0
since: \[\frac{ d^2y }{ dx^2 } = y''\]

Answer 6

You can also use series to solve this. Eulers method.

Answer 7

I am going to assume what I stated above, that y''-xy = 0 and use series
Using: \[y(x) = \sum_{n=0}^{∞}a_nx^n\]
Then I am going to find y' and y''
I am sure that you are able to do these derivatives easily so I am going to leave it to you to figure out the derivatives [NOTE: a and n are both constants]
Next, you are going to plug in y and y'' into the differential [NOTE: make sure to shift down the 2nd differential, you want it as n=0 instead of n=2 to get the series in terms of x^n]

Answer 8

is \[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ dy }{ dx }\frac{ dy }{ dx }\] ???
i got \[y (\ln y-1) =\frac{ x ^{3} }{ 6 }+ c _{i}(x-y) \]

Answer 9

anyway, thx wio n abb0t, for helping me

Answer 10

you should be using power series, the solution is in one of the standard special function
http://en.wikipedia.org/wiki/Airy_function