how to write xy''-(cosx)y'+(sinx)y=0, x0 under the form of [P(x)y']'+[f(x)y]'=0
Please, help

Question

how to write xy''-(cosx)y'+(sinx)y=0, x>0 under the form of [P(x)y']'+[f(x)y]'=0
Please, help

tkhunny · Answer

You could start from the back and give it a good look

Using various rules (product, chain, etc)
[P(x)y']' = P(x)y" + P'(x)y'
[f(x)y]' = f(x)y' + f'(x)y'

loser66 · Answer

[xy']' +[(-1-cos x)y]'=0
[xy']'= [(1+cos x)y]'
integral both sides
xy' =(1+cos x)y
$$y' -\frac{1+cos x}{x}y=0$$
$$\huge y = Ce^{-\int {-\frac{1+cos x}{x}}}$$

loser66 · Answer

not that, ha, I got what I was wrong.

zepdrix · Answer

oh u figured it out? :D

loser66 · Answer

yes, thanks for respond. I got it.

loser66 · Answer

$$y = e^{\mu t}\int Ce^\mu (t)dt +D$$

loser66 · Answer

@ Chillout I am sorry friend. I figured out my mistake and carelessly wrote it down.  it was not that. Actually it is 
$$y = \frac {1}{e^{\mu x dx}}\int C e^{\mu x dx} + D e^{-\ mu x dx}$$ hopefully no more mistypo.

anonymous · Answer

Variation of parameters... what a fun mess! are you trying to solve the equation or just rewrite it bud?

loser66 · Answer

@druminjosh I rewrote it first and then solved it. I knew how to solve it. At that moment, I didn't know how to link the forms. Now, I know it. thanks for respond.