# Homogeneous Equations

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

It can be transformed into a separable equation using the substitution ${y}={u}{t}$ along with the corresponding derivative $\frac{{{d}{y}}}{{{d}{t}}}=\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}$. 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 ${y}'=\frac{{{2}{t}{y}}}{{{{t}}^{{2}}-{{y}}^{{2}}}}$.

This equation is not separable, however it is homogeneous, because ${f{{\left({a}{t},{a}{y}\right)}}}=\frac{{{2}{\left({a}{t}\right)}{\left({a}{y}\right)}}}{{{{\left({a}{t}\right)}}^{{2}}-{{\left({a}{y}\right)}}^{{2}}}}=\frac{{{{a}}^{{2}}\cdot{2}{t}{y}}}{{{{a}}^{{2}}{\left({{t}}^{{2}}-{{y}}^{{2}}\right)}}}=\frac{{{2}{t}{y}}}{{{{t}}^{{2}}-{{y}}^{{2}}}}={f{{\left({t},{y}\right)}}}$.

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

$\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}=\frac{{{2}{t}\cdot{u}{t}}}{{{{t}}^{{2}}-{{\left({u}{t}\right)}}^{{2}}}}$,

Or

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

The last equation can be rewritten as

$\frac{{{{u}}^{{2}}-{1}}}{{{u}{\left({{u}}^{{2}}+{1}\right)}}}{d}{u}=-\frac{{{d}{t}}}{{t}}$.

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

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

Since ${y}={u}{t}$, we have that ${u}=\frac{{y}}{{t}}$ and ${t}{\left({{\left(\frac{{y}}{{t}}\right)}}^{{2}}+{1}\right)}={k}\frac{{y}}{{t}}$, or ${{y}}^{{2}}+{{t}}^{{2}}={k}{y}$.

Now, let's see one more quick example.

Example 2. Solve ${y}'=\frac{{y}}{{{t}+\sqrt{{{t}{y}}}}}$.

This equation is homogeneous, because ${f{{\left({a}{t},{a}{y}\right)}}}=\frac{{{a}{y}}}{{{a}{t}+\sqrt{{{\left({a}{t}\right)}{\left({a}{y}\right)}}}}}=\frac{{y}}{{{t}+\sqrt{{{t}{y}}}}}={f{{\left({t},{y}\right)}}}$.

Using the substitution ${y}={u}{t}$, we obtain that $\frac{{{d}{u}}}{{{d}{t}}}{t}+{u}=\frac{{{u}{t}}}{{{t}+\sqrt{{{t}\cdot{u}{t}}}}}$, which simplifies to $\frac{{{d}{u}}}{{{d}{t}}}{t}=\frac{{u}}{{{1}+\sqrt{{{u}}}}}-{u}$, or $\frac{{{d}{u}}}{{{d}{t}}}{t}=-\frac{{{u}\sqrt{{{u}}}}}{{{1}+\sqrt{{{u}}}}}$: this can be finally rewritten as $\frac{{{1}+\sqrt{{{u}}}}}{{{u}\sqrt{{{u}}}}}{d}{u}=-\frac{{{d}{t}}}{{t}}$.

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

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

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

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