Separable Differential Equations

Consider the differential equation `y'=f(t,y)`, or `(dy)/(dt)=f(t,y)`. If the function `f(t,y)` can be written as the product of the function `g(t)` (function that depends only on `t`) and the function `u(y)` (function that depends only on `y`), such a differential equation is called separable.

Let's see how it is solved.

`(dy)/(dt)=g(t)u(y)`, or `(dy)/(u(y))=g(t)dt`.

Integrating both sides yields `int (dy)/(u(y))=int g(t)dt +C`, where `C` is an arbitrary constant.

Example 1. Solve `(t+1)dt-3/y^4 dy=0`, `y(1)=3`.

This equation is separable and can be rewrritten as `3/y^4 dy=(t+1)dt`.

Integrating both sides yields `int 3/y^4 dy=int (t+1)dt`, or `-1/(y^3)=(t^2)/2+t+C`, where `C` is an arbitrary constant.

Rewriting it a bit, we obtain that `y=-1/root(3)(1/2 t^2+t+C)`. This is the general solution. To find the particular solution, plug the initial values and find the constant `C`.

`y(1)=3=-1/root(3)(1/2 1^2+1+C)-> 3/2+C=-1/27`. So, `C=-83/54`. Thus, the particular solution is ` y=-1/root(3)(1/2 t^2+t-83/54)`.

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

Example 2. Solve `y'=y^2t^2`.

This equation is separable. We can rewrite it as `(dy)/(dt)=y^2t^2`, or `(dy)/y^2=t^2dt`.

Integrating both sides gives `int (dy)/y^2=int t^2dt`, or `-1/y=1/3 t^3+C`, where `C` is an arbitrary constant.

Thus, the general solution is `y=-1/(1/3 t^3+C)`.

There are also 2 particular cases: when `g(t)=1` (the differential equation doesn't contain `t`) or `f(y)=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 `(dy)/(dt)=e^(-y)` or `e^y dy=dt`.

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

So, the general solution is `y=ln(t+C)`.

And one more final example.

Example 4. Solve `y'=sin(t)`, `y(0)=5`.

We can rewrite it as `(dy)/(dt)=sin(t)`, or `dy=sin(t)dt`.

Integrating both sides gives `y=-cos(t)+C`, where `C` is an arbitrary constant.

To find the particular solution, use the initial condition `y(0)=5`:

`y(0)=5=-cos(0)+C ->C=6`

So, the particular solution is `y=-cos(t)+6`.