# 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.