Question

What is a solution to the differential equation `dy/dt=e^t(y-1)^2`?

Answer 1

The General Solution is:

`y = 1-1/(e^t + C)`

Explanation:

We have:

`dy/dt = e^t(y-1)^2`

We can collect terms for similar variables:

`1/(y-1)^2 \ dy/dt = e^t`

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

`int \ 1/(y-1)^2 \ dy = int e^t \ dt`

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

`-1/(y-1) = e^t + C`

And we can readily rearrange for `y`:

`-(y-1) = 1/(e^t + C)`

`:. 1-y = 1/(e^t + C)`

Leading to the General Solution:

`y = 1-1/(e^t + C)`

Answer 2

`y=-1/(e^t+C)+1`

Explanation:

This is a separable differential equation, which means it can be written in the form:

`dy/dx*f(y)=g(x)`

It can be solved by integrating both sides:

`int\ f(y)\ dy=int\ g(x)\ dx`

In our case, we first need to separate the integral into the right form. We can do this by dividing both sides by `(y-1)^2`:

`dy/dt*1/(y-1)^2=e^tcancel((y-1)^2/(y-1)^2)`

`dy/dt*1/(y-1)^2=e^t`

Now we can integrate both sides:

`int\ 1/(y-1)^2\ dy=int\ e^t\ dt`

`int\ 1/(y-1)^2\ dy=e^t+C_1`

We can solve the left hand integral with a substitution of `u=y-1`:

`int\ 1/u^2\ du=e^t+C_1`

`int\ u^-2\ du=e^t+C_1`

`u^-1/(-1)+C_2=e^t+C_1`

Resubstituting (and combining constants) gives:

`-1/(y-1)=e^t+C_3`

Multiply both sides by `y-1`:

`-1=(e^t+C_3)(y-1)`

Divide both sides by `e^t+C_3`:

`-1/(e^t+C_3)=y-1`

`y=-1/(e^t+C)+1`