Solving a Homogeneous ODE: Finding the Solution | Steps and Techniques

[Mechdude]

Homework Statement

this i a problem on homogeneous ode given as an exercise in class
\frac {dy} {dx} = \frac {2xy} {x^2+y^2}
im supposed to find the solution,

Homework Equations

substitute v= \frac {y} {x}
do it as a homogeneous ode and on and on

The Attempt at a Solution

i worked all the way to this point
\int \frac {dv} {dx} = \int \frac {v+v^3} {1+v^2} after the trivial substitution for homogeneous odes and stuff v= \frac {y} {x}
which then i tried separation of variables on it,
\int \frac {1+v^2} {v+v^3}dv = \int \frac {dx} {x}
the right side is trivial so i will not continue to blabber about it,
now separating the left into two equations ;
\int \frac {1} {v+v^3} dv +\int \frac{v^2} {v+v^3} dv =...
the first integral there has me stumped the second well seems doable by substitution, unless its more deceptive than i think since i got an ans of \frac {1} {2} \ln (1+v^2) after doing the substitution u=1+v^2 , who can help me with that first integral \int \frac {1} {v+v^3} dv ?

 

[Hootenanny]

Welcome to Physics Forums.

For your first integral, try taking a factor of v out of the denominator and then using the substitution v = tan(theta).

 

[Mechdude]

thanks man,

 

[HallsofIvy]

Homework Statement

this i a problem on homogeneous ode given as an exercise in class
\frac {dy} {dx} = \frac {2xy} {x+x^3}
im supposed to find the solution,

Homework Equations

substitute v= \frac {y} {x}
do it as a homogeneous ode and on and on

The Attempt at a Solution

i worked all the way to this point
\int \frac {dv} {dx} = \int \frac {v+v^3} {1+v^2} after the trivial substitution for homogeneous odes and stuff v= \frac {y} {x}
which then i tried separation of variables on it,
\int \frac {1+v^2} {v+v^3}dv = \int \frac {dx} {x}
the right side is trivial so i will not continue to blabber about it,
now separating the left into two equations ;
\int \frac {1} {v+v^3} dv +\int \frac{v^2} {v+v^3} dv =...
the first integral there has me stumped the second well seems doable by substitution, unless its more deceptive than i think since i got an ans of \frac {1} {2} \ln (1+v^2) after doing the substitution u=1+v^2 , who can help me with that first integral \int \frac {1} {v+v^3} dv ?

Yes, but why? In addition to being "homogeneous", that de is separable and it is easier doing it that way. In fact, it is probably a good idea to recognize immediately that, for x\ne 0,
\frac{2xy}{x+ x^3}= \frac{2y}{1+ x^2}

So this "separates" as
\frac{dy}{y}= 2\frac{dx}{x^2+ 1}
and both sides are easy to integrate.

 

[Mechdude]

Yes, but why? In addition to being "homogeneous", that de is separable and it is easier doing it that way. In fact, it is probably a good idea to recognize immediately that, for x\ne 0,
\frac{2xy}{x+ x^3}= \frac{2y}{1+ x^2}

So this "separates" as
\frac{dy}{y}= 2\frac{dx}{x^2+ 1}
and both sides are easy to integrate.

you ask why?
because its a class exercise for the method :) , and i realize i wrote the wrong question but the working for the right question so here is the correct question \frac{dx}{dy}=\frac{2xy} {x^2+y^2} i will correct the posting too

 

[Mechdude]

Welcome to Physics Forums.




For your first integral, try taking a factor of v out of the denominator and then using the substitution v = tan(theta).

is this what you mean? : \frac{1}{v(1+v^2)} and substituting v=tan(\theta)
dv= tan^2(\theta) + 1 now we have \int \frac{tan^2(\theta) + 1 }{tan\theta (1 + tan^2 (\theta))} d\theta after cancelling :
\int \frac {d\theta} {tan(\theta)} but tan is identical to \frac{sin}{cos} using u=sin(\theta) we got now: \int \frac{du} {u} and the grand finale \ln(sin(\theta))

 

[Hootenanny]

is this what you mean? : \frac{1}{v(1+v^2)} and substituting v=tan(\theta)
dv= tan^2(\theta) + 1 now we have \int \frac{tan^2(\theta) + 1 }{tan\theta (1 + tan^2 (\theta))} d\theta after cancelling :
\int \frac {d\theta} {tan(\theta)} but tan is identical to \frac{sin}{cos} using u=sin(\theta) we got now: \int \frac{du} {u} and the grand finale \ln(sin(\theta))

That is indeed what I mean. However, you should note that this integral no longer applies to your "corrected" question.

 

[Mechdude]

That is indeed what I mean. However, you should note that this integral no longer applies to your "corrected" question.

why? The working i gave was for the corrected question , not the first erroneous one

 

[Hootenanny]

why? The working i gave was for the corrected question , not the first erroneous one

Starting from your corrected ODE,

\frac{dy}{dx} = \frac{2xy}{x^2 + y^2}

and substituting y=vx yields,

x\frac{dv}{dx} = \frac{2v}{1+v^2} - v

x\frac{dv}{dx} = \frac{v - v^3}{1+v^2}

\int\frac{1+v^2}{v\left(1-v^2\right)}\;dv = \int\frac{dx}{x}

Which does not match your integrals. It seems that you have dropped a sign.

 

[Mechdude]

Starting from your corrected ODE,




\frac{dy}{dx} = \frac{2xy}{x^2 + y^2}


and substituting y=vx yields,

x\frac{dv}{dx} = \frac{2v}{1+v^2} - v

x\frac{dv}{dx} = \frac{v - v^3}{1+v^2}

\int\frac{1+v^2}{v\left(1-v^2\right)}\;dv = \int\frac{dx}{x}

Which does not match your integrals. It seems that you have dropped a sign.

thanks, my signs need some more work,
from here is this how this would be done: \int\frac{1}{v\left(1-v^2\right)}\;dv
let v= sin\theta , dv= cos\theta d\theta and we have,
\int \frac {cos\theta}{sin\theta cos^2 \theta }d\theta
eventually
2\int \frac {1}{sin2\theta}d\theta and
2\int csc2\theta d\theta

 

[Hootenanny]

From here

2\int \frac {1}{sin2\theta}d\theta

note that

\sin2\theta = \frac{2\tan\theta}{1+tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta}

Hence,

2\int \frac {1}{sin2\theta}d\theta = \int\frac{\sec^2\theta}{\tan\theta}d\theta

And note that,

\frac{d}{d\theta}\tan\theta = \sec^2\theta

From here, your integral is trivial.

 

[ehild]

<br /> \frac{1+v^2}{v\left(1-v^2\right)}\ = \frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}<br />

ehild

 

[Mechdude]

Thanks, i get the method now