# 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.