1st order O.D.E.: y'-2y/x=-6y^2

[Joki]

Hi there guys,

Just wondering the best way to do about solving the equation;
y'-2y/x=-6y^2 is?
I would normally go integrating factor but the right hand side is in terms of y and not x.
Alternatively my algebra is not the greatest and i'm having a hard time separating the equation out.
Any help would be much appreciated

 

[stapel]

Just wondering the best way to do about solving the equation;
y'-2y/x=-6y^2 is?

As posted, the equation is this:

. . . . .\(\displaystyle y'\, -\, \dfrac{2y}{x}\, =\, -6y^2\)

Is this what you meant, or was the left-hand side supposed to have grouping symbols like this? (y' - 2y)/x

. . . . .\(\displaystyle \dfrac{y'\, -\, 2y}{x}\, =\, -6y^2\)

Thank you!

 

[Joki]

Apologies stapel,

The equation is identical to the first one with the terms separated and not all being divided by x.

Btw thanks for the quick reply :grin:

 

[Deleted member 4993]

Hi there guys,

Just wondering the best way to do about solving the equation;
y'-2y/x=-6y^2 is?
I would normally go integrating factor but the right hand side is in terms of y and not x.
Alternatively my algebra is not the greatest and i'm having a hard time separating the equation out.
Any help would be much appreciated

y' + 6y² = 2y/x

y'/y + 6y = 2/x

w' + 6e^w = 2/x where ln(y) = w

Now continue....

 

[Joki]

y' + 6y² = 2y/x

y'/y + 6y = 2/x

w' + 6e^w = 2/x where ln(y) = w

Now continue....

Just to clarify is there more than one way of solving this as I can choose from variable separation, 1st order homogeneous [dy/dx=f(y/x)], exact eqns and linear eqns (Bernoulli and integration factor etc.). I have ruled out exact eqns and dont think its linear eqns so that just leaves first order homogeneous and variable sep.

Cheers

Hang on I think that the Bernoulli may do it

 

[Joki]

So just to check I think I have the answer I just hope to god it's correct!

y=x/(6x²+C)

 

[Deleted member 4993]

So just to check I think I have the answer I just hope to god it's correct!

y=x/(6x²+C)

For checking correctness - apply your solution to the given differential equation.