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Homogeneous Equations

If in the differential equation y′=f(t,y){y}'={f{{\left({t},{y}\right)}}}, the function f(t,y){f{{\left({t},{y}\right)}}} has the property that f(at,ay)=f(t,y){f{{\left({a}{t},{a}{y}\right)}}}={f{{\left({t},{y}\right)}}}, such a differential equation is called homogeneous.

It can be transformed into a separable equation using the substitution y=ut{y}={u}{t} along with the corresponding derivative dydt=dudtt+u\frac{{{d}{y}}}{{{d}{t}}}=\frac{{{d}{u}}}{{{d}{t}}}{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′=2tyt2−y2{y}'=\frac{{{2}{t}{y}}}{{{{t}}^{{2}}-{{y}}^{{2}}}}.

This equation is not separable, however it is homogeneous, because f(at,ay)=2(at)(ay)(at)2−(ay)2=a2⋅2tya2(t2−y2)=2tyt2−y2=f(t,y){f{{\left({a}{t},{a}{y}\right)}}}=\frac{{{2}{\left({a}{t}\right)}{\left({a}{y}\right)}}}{{{{\left({a}{t}\right)}}^{{2}}-{{\left({a}{y}\right)}}^{{2}}}}=\frac{{{{a}}^{{2}}\cdot{2}{t}{y}}}{{{{a}}^{{2}}{\left({{t}}^{{2}}-{{y}}^{{2}}\right)}}}=\frac{{{2}{t}{y}}}{{{{t}}^{{2}}-{{y}}^{{2}}}}={f{{\left({t},{y}\right)}}}.

Using the substitution y=ut{y}={u}{t} and the corresponding derivative dydt=dudtt+u\frac{{{d}{y}}}{{{d}{t}}}=\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}, we obtain that

dudtt+u=2t⋅utt2−(ut)2\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}=\frac{{{2}{t}\cdot{u}{t}}}{{{{t}}^{{2}}-{{\left({u}{t}\right)}}^{{2}}}},

Or

dudtt+u=2u1−u2\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}=\frac{{{2}{u}}}{{{1}-{{u}}^{{2}}}}, which gives dudtt=−u(u2+1)u2−1\frac{{{d}{u}}}{{{d}{t}}}{t}=-\frac{{{u}{\left({{u}}^{{2}}+{1}\right)}}}{{{{u}}^{{2}}-{1}}}.

The last equation can be rewritten as

u2−1u(u2+1)du=−dtt\frac{{{{u}}^{{2}}-{1}}}{{{u}{\left({{u}}^{{2}}+{1}\right)}}}{d}{u}=-\frac{{{d}{t}}}{{t}}.

Using partial fraction decomposition, we obtain that (2uu2+1−1u)du=−dtt{\left(\frac{{{2}{u}}}{{{{u}}^{{2}}+{1}}}-\frac{{1}}{{u}}\right)}{d}{u}=-\frac{{{d}{t}}}{{t}}.

Integrating both sides gives ln⁡(u2+1)−ln⁡(∣u∣)=−ln⁡(∣t∣)+C{\ln{{\left({{u}}^{{2}}+{1}\right)}}}-{\ln{{\left({\left|{u}\right|}\right)}}}=-{\ln{{\left({\left|{t}\right|}\right)}}}+{C}, or t(u2+1)=ku{t}{\left({{u}}^{{2}}+{1}\right)}={k}{u} (C=ln⁡(∣k∣))\left({C}={\ln{{\left({\left|{k}\right|}\right)}}}\right).

Since y=ut{y}={u}{t}, we have that u=yt{u}=\frac{{y}}{{t}} and t((yt)2+1)=kyt{t}{\left({{\left(\frac{{y}}{{t}}\right)}}^{{2}}+{1}\right)}={k}\frac{{y}}{{t}}, or y2+t2=ky{{y}}^{{2}}+{{t}}^{{2}}={k}{y}.

Now, let's see one more quick example.

Example 2. Solve y′=yt+ty{y}'=\frac{{y}}{{{t}+\sqrt{{{t}{y}}}}}.

This equation is homogeneous, because f(at,ay)=ayat+(at)(ay)=yt+ty=f(t,y){f{{\left({a}{t},{a}{y}\right)}}}=\frac{{{a}{y}}}{{{a}{t}+\sqrt{{{\left({a}{t}\right)}{\left({a}{y}\right)}}}}}=\frac{{y}}{{{t}+\sqrt{{{t}{y}}}}}={f{{\left({t},{y}\right)}}}.

Using the substitution y=ut{y}={u}{t}, we obtain that dudtt+u=utt+t⋅ut\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}=\frac{{{u}{t}}}{{{t}+\sqrt{{{t}\cdot{u}{t}}}}}, which simplifies to dudtt=u1+u−u\frac{{{d}{u}}}{{{d}{t}}}{t}=\frac{{u}}{{{1}+\sqrt{{{u}}}}}-{u}, or dudtt=−uu1+u\frac{{{d}{u}}}{{{d}{t}}}{t}=-\frac{{{u}\sqrt{{{u}}}}}{{{1}+\sqrt{{{u}}}}}: this can be finally rewritten as 1+uuudu=−dtt\frac{{{1}+\sqrt{{{u}}}}}{{{u}\sqrt{{{u}}}}}{d}{u}=-\frac{{{d}{t}}}{{t}}.

Furhter rewriting the left side a bit gives (1u32+1u)du=−dtt{\left(\frac{{1}}{{{{u}}^{{\frac{{3}}{{2}}}}}}+\frac{{1}}{{u}}\right)}{d}{u}=-\frac{{{d}{t}}}{{t}}.

Integrating both sides yields −2u+ln⁡(∣u∣)=−ln⁡(∣t∣)+C-\frac{{2}}{\sqrt{{{u}}}}+{\ln{{\left({\left|{u}\right|}\right)}}}=-{\ln{{\left({\left|{t}\right|}\right)}}}+{C}.

Using back substitution, we obtain the result: −2yt+ln⁡(∣yt∣)=−ln⁡(∣t∣)+C-\frac{{2}}{{\sqrt{{\frac{{y}}{{t}}}}}}+{\ln{{\left({\left|\frac{{y}}{{t}}\right|}\right)}}}=-{\ln{{\left({\left|{t}\right|}\right)}}}+{C}, which can be simplified to −2ty+ln⁡(∣y∣)=C-{2}\sqrt{{\frac{{t}}{{y}}}}+{\ln{{\left({\left|{y}\right|}\right)}}}={C}.

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