# Homogeneous Equations

If in differential equation y'=f(t,y), function f(t,y) has the property that f(at,ay)=f(t,y) then such differential equation is called homogeneous.

It can be transformed into separable equation using substitution ` y=ut` along with the corresponding derivative `(dy)/(dt)=(du)/(dt)t+u`. Resulting equation is solved as separable equation and required solution is obtained by back subtitution.

Let's work a couple of examples.

Example 1. Solve `y'=(2ty)/(t^2-y^2)` .

This equation is not separable, however it is homogeneous, because `f(at,ay)=(2(at)(ay))/((at)^2-(ay)^2)=(a^2*2ty)/(a^2(t^2-y^2))=(2ty)/(t^2-y^2)=f(t,y)`.

Using the substitution `y=ut` and the corresponding derivative `(dy)/(dt)=(du)/(dt)t+u` we obtain that

`(du)/(dt)t+u=(2t*ut)/(t^2-(ut)^2)`

Or

` (du)/(dt)t+u=(2u)/(1-u^2)` , which gives `(du)/(dt)t=-(u(u^2+1))/(u^2-1)`

Last equation can be rewritten as

`(u^2-1)/(u(u^2+1))du=-(dt)/t`

Using partial fraction we obtain that `((2u)/(u^2+1)-1/u)du=-(dt)/t` .

Integrating both sides gives `ln(u^2+1)-ln(|u|)=-ln(|t|)+C` or `t(u^2+1)=ku` (C=ln(|k|)).

Since y=ut then `u=y/t` and we have that `t((y/t)^2+1)=ky/t` or `y^2+t^2=ky`.

Example 2. Solve `y'=y/(t+sqrt(ty))`

This equation is homogeneous because `f(at,ay)=(ay)/(at+sqrt((at)(ay)))=y/(t+sqrt(ty))=f(t,y)`.

Using substitution `y=ut` we obtaint that `(du)/(dt)t+u=(ut)/(t+sqrt(t*ut))` which simplifies to `(du)/(dt)t=u/(1+sqrt(u))-u` or `(du)/(dt)t=-(u sqrt(u))/(1+sqrt(u))` that can be finally rewritten as `(1+sqrt(u))/(usqrt(u))du=-(dt)/t`

Furhter rewriting left part a bit gives `(1/(u^(3/2))+1/u)du=-(dt)/t` .

Integrating both sides yields `-2/sqrt(u)+ln(|u|)=-ln(|t|)+C`.

Using back substitution we obtain result: `-2/(sqrt(y/t))+ln(|y/t|)=-ln(|t|)+C`, which can be simplified to `-2sqrt(t/y)+ln(|y|)=C`

Note that we can't find explicit solution, so we leave it as implicit solution.