Differential equations Problem?

[tdabboud]

Need Help with a couple diff eq problems:

2.) show y=cx^2 - x is a solution of xy' - 2y + x =0

I was trying to separate the variables so i could integrate but i cannot get it to work out.
I have tried so many things from adding 2y to both sides and and multiply by the inverse dx/dy. but no avail

3.) use Integrating factor to solve: xdy - ydx = x^2ydy
I subtracted x^2ydy to left side: and used this value as p(x) for the Int.Fact. but I do not think this is correct either. I cannot figure out what to use as p(x)

dy/dx + p(x)y = f(x)

 

[rock.freak667]

Need Help with a couple diff eq problems:

2.) show y=cx^2 - x is a solution of xy' - 2y + x =0

I was trying to separate the variables so i could integrate but i cannot get it to work out.
I have tried so many things from adding 2y to both sides and and multiply by the inverse dx/dy. but no avail

There is an easier way is that if you are given a solution 'y', it will satisfy the DE. So just find y', then sub it into the DE, if you get 0, then y=cx² is a solution.

3.) use Integrating factor to solve: xdy - ydx = x^2ydy
I subtracted x^2ydy to left side: and used this value as p(x) for the Int.Fact. but I do not think this is correct either. I cannot figure out what to use as p(x)

dy/dx + p(x)y = f(x)

Your equation will not be in that form, instead what you need to do is write it in the form

N dy + M dx = 0

If (∂N/∂x - ∂M/∂y)/M = f(x) then e^{∫-f(x) dx} is an integrating factor

If (∂N/∂x - ∂M/∂y)/N = g(y) then e^{∫g(y) dy} is an integrating factor.

Unfortunately I cannot remember if my signs for the integrating factors are correct, so I suggest you look them up for the correct thing. The general form is similar, but may differ by a + or - sign.