Larrytsai said:The Attempt at a Solution
I divide x^2 on both sides and end up with...
y'= y^2/x^2 + y/x
then using the hint, i get
y' = u^2 + u
but i do not know how to solve the DE from here.
rock.freak667 said:You should have
(1/x2)y'2 =u2+u
if u=y/x or y=ux, what is y' ?
Larrytsai said:hmm how did u get y'/x^2?
EDIT:
Im a little confused, when taking the derivative of y= ux, am i differentiating w.r.t x?
rock.freak667 said:yes, you differentiate with respect to x.
The left side was y'2, so dividing both sides by x2 means that the left side becomes y'2/x2
Larrytsai said:ohh sorry, only the x was squared for the left term not the y'
Larrytsai said:k so taking the derivative of y=ux, i have y' = xu' + u, and i just sub it into the equation, then i have xu' = u^2 + u which I can simply put it in the general first oder ODE form, and solve by separation
ohh yah, I meant y' * x^2rock.freak667 said:I have small query, is the term on the left yx2 OR (y')2 ?
Because y(x) is a function in x and y'(x) is its derivative, so y'(x)2 is the derivative squared, so which is it?
Larrytsai said:ohh yah, I meant y' * x^2
Larrytsai said:k so taking the derivative of y=ux, i have y' = xu' + u, and i just sub it into the equation, then i have xu' = u^2 + u which I can simply put it in the general first oder ODE form, and solve by separation
rock.freak667 said:ah in that case, this post under this one is correct.
You will get xu' + u = u2+u, the 'u' will cancel and then you can solve by separation. then you need to remember that you will get 'u' and not 'y', but you know that y=ux.
To solve this ODE, we first substitute the given hint, u = y/x, into the equation. This results in a new equation with only u and its derivatives. From there, we can use standard techniques such as separation of variables or integrating factors to solve for u. Once we have found the solution for u, we can then substitute back in y/x for u to obtain the solution for y.
The hint u = y/x allows us to transform the original equation into a separable form, making it easier to solve. Additionally, it helps us to identify a new dependent variable, u, which can simplify the overall solution process.
Yes, there are other substitution methods that can be used to solve this ODE. Some other common substitutions include v = y^2, w = y'/y, or z = y - x. However, the given hint u = y/x is specifically chosen to make the solution process more efficient and straightforward.
Yes, since we are dealing with derivatives, the ODE may have restrictions on the values of x and y in order for the equation to be valid. For example, the equation may not be defined for certain values of x and y, or it may have a restricted domain for the solution to exist.
The given hint, u = y/x, is specifically chosen for this particular ODE. It may not be applicable to other types of first order ODEs. However, similar substitution methods can be used to transform other ODEs into a more manageable form for solving.