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

2 Answers
Apr 4, 2018

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) #

Apr 4, 2018

#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#