Homogeneous Equations

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 $$${{y}}^{{2}}+{{t}}^{{2}}={k}{y}$$$.

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.