Solving a First Order ODE with y'(x)^2 = y^2 + xy and the Hint: u = y/x

[Larrytsai]

Homework Statement

y'(x)^2 = y^2 +xy

Homework Equations

hint : let u = y/x

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]

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.

You should have

(1/x²)y'² =u²+u

if u=y/x or y=ux, what is y' ?

 

[viscousflow]

whoops just gave the answer, don't want to destroy your thunder rock.

 

[Larrytsai]

You should have

(1/x²)y'² =u²+u

if u=y/x or y=ux, what is y' ?

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]

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?

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]

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

ohh sorry, only the x was squared for the left term not the y'

 

[Larrytsai]

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]

ohh sorry, only the x was squared for the left term not the y'

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

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]

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?

ohh yah, I meant y' * x^2

 

[rock.freak667]

ohh yah, I meant y' * x^2

ah in that case, this post under this one is correct.

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

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.

 

[Larrytsai]

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.

great! thanks a lot for your help