# Separable Differential Equations

Consider the differential equation ${y}'={f{{\left({t},{y}\right)}}}$, or $\frac{{{d}{y}}}{{{d}{t}}}={f{{\left({t},{y}\right)}}}$.

If the function ${f{{\left({t},{y}\right)}}}$ can be written as the product of the function ${g{{\left({t}\right)}}}$ (function that depends only on ${t}$) and the function ${u}{\left({y}\right)}$ (function that depends only on ${y}$), such a differential equation is called separable.

Let's see how it is solved.

$\frac{{{d}{y}}}{{{d}{t}}}={g{{\left({t}\right)}}}{u}{\left({y}\right)}$, or $\frac{{{d}{y}}}{{{u}{\left({y}\right)}}}={g{{\left({t}\right)}}}{d}{t}$.

Integrating both sides yields $\int\frac{{{d}{y}}}{{{u}{\left({y}\right)}}}=\int{g{{\left({t}\right)}}}{d}{t}+{C}$, where ${C}$ is an arbitrary constant.

Example 1. Solve ${\left({t}+{1}\right)}{d}{t}-\frac{{3}}{{{y}}^{{4}}}{d}{y}={0}$, ${y}{\left({1}\right)}={3}$.

This equation is separable and can be rewrritten as $\frac{{3}}{{{y}}^{{4}}}{d}{y}={\left({t}+{1}\right)}{d}{t}$.

Integrating both sides yields $\int\frac{{3}}{{{y}}^{{4}}}{d}{y}=\int{\left({t}+{1}\right)}{d}{t}$, or $-\frac{{1}}{{{{y}}^{{3}}}}=\frac{{{{t}}^{{2}}}}{{2}}+{t}+{C}$, where ${C}$ is an arbitrary constant.

Rewriting it a bit, we obtain that ${y}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{t}}^{{2}}+{t}+{C}}}}}$. This is the general solution. To find the particular solution, plug in the initial values and find the constant ${C}$.

${y}{\left({1}\right)}={3}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{1}}^{{2}}+{1}+{C}}}}}\to\frac{{3}}{{2}}+{C}=-\frac{{1}}{{27}}$. So, ${C}=-\frac{{83}}{{54}}$. Thus, the particular solution is ${y}=-\frac{{1}}{{\sqrt[{{3}}]{{\frac{{1}}{{2}}{{t}}^{{2}}+{t}-\frac{{83}}{{54}}}}}}$.

Now, let's take a look at another example.

Example 2. Solve ${y}'={{y}}^{{2}}{{t}}^{{2}}$.

This equation is separable. We can rewrite it as $\frac{{{d}{y}}}{{{d}{t}}}={{y}}^{{2}}{{t}}^{{2}}$, or $\frac{{{d}{y}}}{{{y}}^{{2}}}={{t}}^{{2}}{d}{t}$.

Integrating both sides gives $\int\frac{{{d}{y}}}{{{y}}^{{2}}}=\int{{t}}^{{2}}{d}{t}$, or $-\frac{{1}}{{y}}=\frac{{1}}{{3}}{{t}}^{{3}}+{C}$, where ${C}$ is an arbitrary constant.

Thus, the general solution is ${y}=-\frac{{1}}{{\frac{{1}}{{3}}{{t}}^{{3}}+{C}}}$.

There are also 2 particular cases: when ${g{{\left({t}\right)}}}={1}$ (the differential equation doesn't contain ${t}$) or ${f{{\left({y}\right)}}}={1}$ (the differential equation doesn't contain ${y}$).

Let's do some more practice.

Example 3. Solve ${y}'={{e}}^{{-{y}}}$.

We can rewrite it as $\frac{{{d}{y}}}{{{d}{t}}}={{e}}^{{-{y}}}$, or ${{e}}^{{y}}{d}{y}={d}{t}$.

Integrating both sides gives ${{e}}^{{y}}={t}+{C}$, where ${C}$ is an arbitrary constant.

So, the general solution is ${y}={\ln{{\left({t}+{C}\right)}}}$.

And one more final example.

Example 4. Solve ${y}'={\sin{{\left({t}\right)}}}$, ${y}{\left({0}\right)}={5}$.

We can rewrite it as $\frac{{{d}{y}}}{{{d}{t}}}={\sin{{\left({t}\right)}}}$, or ${d}{y}={\sin{{\left({t}\right)}}}{d}{t}$.

Integrating both sides gives ${y}=-{\cos{{\left({t}\right)}}}+{C}$, where ${C}$ is an arbitrary constant.

To find the particular solution, use the initial condition ${y}{\left({0}\right)}={5}$:

${y}{\left({0}\right)}={5}=-{\cos{{\left({0}\right)}}}+{C}\to{C}={6}$

So, the particular solution is ${y}=-{\cos{{\left({t}\right)}}}+{6}$.