Homogeneous Equations

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

It can be transformed into a separable equation using the substitution ` y=ut` along with the corresponding derivative `(dy)/(dt)=(du)/(dt)t+u`. The resulting equation is solved as a separable equation, and the 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)`.

The last equation can be rewritten as

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

Using partial fraction decomposition, 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`, we have that `u=y/t` and `t((y/t)^2+1)=ky/t`, or `y^2+t^2=ky`.

Now, let's see one more quick example.

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 the substitution `y=ut`, we obtain 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))`: this can be finally rewritten as `(1+sqrt(u))/(usqrt(u))du=-(dt)/t`.

Furhter rewriting the left side 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 the 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 an explicit solution, so we leave it as an implicit solution.