Separable Differential Equations

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

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

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

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 arbitrary constant.

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

There are also 2 particular cases: when `g(t)=1` (differential equation doesn't contain t) or `f(y)=1` (differential equation doesn't contain y).

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 arbitrary constant.

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

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 arbitrary constant.

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

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

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