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

Question

Emilia Aguilar · Accepted Answer

The General Solution is:

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

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) #

Cameron Alexander · Answer

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

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#

Ezekiel Alexander · Answer

undefined The solution to the given differential equation $ \frac{dy}{dt} = e^t(y-1)^2 $ is $ y(t) = 1 + \frac{1}{Ce^{-t} - 1} $, where $ C $ is an arbitrary constant.