Not to be confused with homogeneous equations, an equation homogeneous in x and y of degree n is an equation of the form
Such that
.
Then the equation can take the form
Which is essentially another in the form
.
If we can solve this equation for
, then we can easily use the substitution method mentioned earlier to solve this equation. Suppose, however, that it is more easily solved for
,
So that
.
We can differentiate this to get
Then re-arranging things,
So that upon integrating,
We get
Thus, if we can eliminate y' between two simultaneous equations
and
,
then we can obtain the general solution..
A function P is homogeneous of order
if
. A homogeneous ordinary differential equation is an equation of the form P(x,y)dx+Q(x,y)dy=0 where P and Q are homogeneous of the same order.
The first usage of the following method for solving homogeneous ordinary differential equations was by Leibniz in 1691. Using the substitution y=vx or x=vy, we can make turn the equation into a separable equation.
![{\displaystyle {\frac {dy}{dx}}=F\left({\frac {y}{x}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60d8bae215d979a3cb68ba3ef2388195006edf08)
![{\displaystyle v(x,y)={\frac {y}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/526099c429ac97cd624603dd0e900e3e9f71b514)
![{\displaystyle y=vx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0667cc7b59011f76322626a490fed212bc1a72b)
Now we need to find v':
![{\displaystyle {\frac {dy}{dx}}=v+{\frac {dv}{dx}}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1891cbe11d07e283720b87df1c3f2210d0ced23b)
Plug back into the original equation
![{\displaystyle v+x{\frac {dv}{dx}}=F(v)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a9d524c29700f688888f96c972d5b1e4ae82b75)
![{\displaystyle {\frac {dv}{dx}}={\frac {F(v)-v}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e64e83416f10cf6c4ee492b96b522d65d0dd5bc)
- Solve for v(x), then plug into the equation of v to get y
![{\displaystyle y(x)=xv(x)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cbec01f561764c7e3eb7069e4a34273da2de89c)
Again, don't memorize the equation. Remember the general method, and apply it.
![{\displaystyle {\frac {dy}{dx}}=5{\frac {y}{x}}+3{\frac {x}{y}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b08a3144994802004c6ab83e653520687732972)
Let's use
. Solve for
![{\displaystyle y=vx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0667cc7b59011f76322626a490fed212bc1a72b)
![{\displaystyle {\frac {dy}{dx}}=v+x{\frac {dv}{dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e61e0f57aea6b9c8bd7be9daa6dcebce36c04bce)
Now plug into the original equation
![{\displaystyle v+x{\frac {dv}{dx}}=5v+{\frac {3}{v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ef8c71a5d7069943e819992a44d10768be6ad97)
![{\displaystyle x{\frac {dv}{dx}}=4v+{\frac {3}{v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49568567ef2343cf98d44de29035498feadf7496)
![{\displaystyle v{\frac {dv}{dx}}={\frac {(4v^{2}+3)}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e3cbc218d872b1dad2b8ac815b6fd3629979f81)
Solve for v
![{\displaystyle {\frac {vdv}{4v^{2}+3}}={\frac {dx}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8de0f2b8092ce3a1d0a34f79ac9a72089c67c452)
![{\displaystyle \int {\frac {vdv}{4v^{2}+3}}=\int {\frac {dx}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/387408418e363ec76b60de4efaebd3a6e964a6e4)
![{\displaystyle {\frac {1}{8}}\ln(4v^{2}+3)=\ln(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ef77f6ac023ec7b59a9eb7ae3e7f0ea111ba2bc)
![{\displaystyle 4v^{2}+3=e^{8\ln(x)}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a30a18c2ff2a89e756c1b31eb0557aa1b4dc9fd2)
![{\displaystyle 4v^{2}+3=e^{\ln(x^{8})}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86fdd344242f7103607bf99895dd84a14d09be09)
![{\displaystyle 4v^{2}+3=x^{8}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1d250fff007d6ba5040014da0e9ed36dbe21a8f)
![{\displaystyle v^{2}={\frac {x^{8}-3}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf95fe04e28c11e58f54cb70cba9128b00e490bf)
Plug into the definition of v to get y.
![{\displaystyle y=vx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0667cc7b59011f76322626a490fed212bc1a72b)
![{\displaystyle y^{2}=v^{2}x^{2}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b545ab458a5f11cff99ebd08e294b092b8fe6861)
![{\displaystyle y^{2}={\frac {x^{10}-3x^{2}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/400a0d57569a7eff2f4a7492faa192e8c6b81c94)
We leave it in
form, since solving for y would lose information.
Note that there should be a constant of integration in the general solution. Adding it is left as an exercise.
![{\displaystyle {\frac {dy}{dx}}={\frac {x}{\sin({\frac {y}{x}})}}+{\frac {y}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3363e0fc1334e6f03271c702c6fe92e211e290c3)
Lets use
again. Solve for
![{\displaystyle y=vx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0667cc7b59011f76322626a490fed212bc1a72b)
![{\displaystyle {\frac {dy}{dx}}=v+x{\frac {dv}{dx}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e61e0f57aea6b9c8bd7be9daa6dcebce36c04bce)
Now plug into the original equation
![{\displaystyle v+x{\frac {dv}{dx}}={\frac {x}{\sin(v)}}+v}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8e8593590d58730aec5c301390f9b630bf237b6)
![{\displaystyle x{\frac {dv}{dx}}={\frac {x}{\sin(v)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0aa7e482814efb36aa17c326a7ca819f4c55ca5)
![{\displaystyle \sin(v)dv=dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86b0c0d7ab7bbde83b6f8d1bf6a165379099c3cb)
Solve for v:
![{\displaystyle \int \sin(v)dv=\int dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc8a7d9e55d7625c75cae15d903786cff44cabe7)
![{\displaystyle -\cos v=x+C\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc4a0cc54c34b3d55ef8ededa6bec304afd0b04c)
![{\displaystyle v=\arccos(-x+C)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/933f55474324d0f7184361a65fc3bc8fa022f9ea)
Use the definition of v to solve for y.
![{\displaystyle y=vx\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0667cc7b59011f76322626a490fed212bc1a72b)
![{\displaystyle y=\arccos(-x+C)x\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce15c093723c46f2dea0525624c5a3b9ef0ad726)
An equation that is a function of a quotient of linear expressions
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Given the equation
,
We can make the substitution x=x'+h and y=y'+k where h and k satisfy the system of linear equations:
![{\displaystyle a_{1}h+b_{1}k+c_{1}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed13520618ad2272550a1048383ebcc24a34e1fb)
Which turns it into a homogeneous equation of degree 0: