# 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.