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Separable Differential Equations

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

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

Let's see how it is solved.

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

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

Example 1. Solve (t+1)dt−3y4dy=0{\left({t}+{1}\right)}{d}{t}-\frac{{3}}{{{y}}^{{4}}}{d}{y}={0}, y(1)=3{y}{\left({1}\right)}={3}.

This equation is separable and can be rewrritten as 3y4dy=(t+1)dt\frac{{3}}{{{y}}^{{4}}}{d}{y}={\left({t}+{1}\right)}{d}{t}.

Integrating both sides yields ∫3y4dy=∫(t+1)dt\int\frac{{3}}{{{y}}^{{4}}}{d}{y}=\int{\left({t}+{1}\right)}{d}{t}, or −1y3=t22+t+C-\frac{{1}}{{{{y}}^{{3}}}}=\frac{{{{t}}^{{2}}}}{{2}}+{t}+{C}, where C{C} is an arbitrary constant.

Rewriting it a bit, we obtain that y=−112t2+t+C3{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{C}.

y(1)=3=−11212+1+C3→32+C=−127{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=−8354{C}=-\frac{{83}}{{54}}. Thus, the particular solution is y=−112t2+t−83543{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′=y2t2{y}'={{y}}^{{2}}{{t}}^{{2}}.

This equation is separable. We can rewrite it as dydt=y2t2\frac{{{d}{y}}}{{{d}{t}}}={{y}}^{{2}}{{t}}^{{2}}, or dyy2=t2dt\frac{{{d}{y}}}{{{y}}^{{2}}}={{t}}^{{2}}{d}{t}.

Integrating both sides gives ∫dyy2=∫t2dt\int\frac{{{d}{y}}}{{{y}}^{{2}}}=\int{{t}}^{{2}}{d}{t}, or −1y=13t3+C-\frac{{1}}{{y}}=\frac{{1}}{{3}}{{t}}^{{3}}+{C}, where C{C} is an arbitrary constant.

Thus, the general solution is y=−113t3+C{y}=-\frac{{1}}{{\frac{{1}}{{3}}{{t}}^{{3}}+{C}}}.

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

Let's do some more practice.

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

We can rewrite it as dydt=e−y\frac{{{d}{y}}}{{{d}{t}}}={{e}}^{{-{y}}}, or eydy=dt{{e}}^{{y}}{d}{y}={d}{t}.

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

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

And one more final example.

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

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

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

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

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

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