What is the solution of the differential equation # dy/dx = 2y(5-3y) # with #y(0)=2#?

Question

What is the solution of the differential equation # dy/dx = 2y(5-3y) # with  #y(0)=2#?

Scarlett Allison · Accepted Answer

# y = (10e^(10x))/(6e^(10x)-1) #

We have: # dy/dx = 2y(5-3y) # We can "separate the variables" in this First Order Separable ODE to obtain # int \ 1/(y(5-3y)) \ dy = int \ 2 \ dx # By breaking down the integrand into partial fractions, it is possible to integrate the LHS while the RHS remains trivial. # 1/(y(5-3y)) -= A/y+ B/(5-3y) # # " " = (A(5-3y)+By)/(y(5-3y)) # Getting to the answer: # 1 -= A(5-3y)+By # Where #A,B# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method: Put # y =0 => 1=5A => A=1/5# Put # x = 5/3 => 1 = (5B)/3 => B = 3/5 # Thus, we can now write: # \ \ \ \ \ \ \ \ \int \ (1/5)/y + (3/5)/(5-3y) \ dy = int \ 2 \ dx # # :. 1/5 int \ 1/y - 3/(3y-5) \ dy = int \ 2 \ dx # And after integrating, we obtain: # 1/5 { ln|y| - ln|3y-5|} = 2x + C # Using the initial condition #y(0)=2# then: # 1/5 { ln2 - ln1} = C => C= 1/5ln 2# Thus, we have: # 1/5 { ln|y| - ln|3y-5|} = 2x + 1/5ln 2 # # :. ln|y| - ln|3y-5| = 10x + ln 2 # # :. ln|y/(3y-5)| = 10x + ln 2 # # :. y/(3y-5) = e^(10x + ln 2) # # :. y = (3y-5) 2e^(10x) # # :. y = 6ye^(10x) - 10e^(10x) # # :. 6ye^(10x) - y = 10e^(10x) # # :. (6e^(10x)-1)y = 10e^(10x) # # :. y = (10e^(10x))/(6e^(10x)-1) #

Elijah Abram · Answer

#y(x) = 10/(6-e^(-10x))#

This differential equation can be separated:

#dy/dx = 2y(5-3y)#

#dy/(2y(5-3y)) = dx#

#(1) " " int dy/(2y(5-3y)) = int dx#

Solve the integral in #y# using partial fractions:

#1/(2y(5-3y)) = A/(2y) + B/(5-3y)#

#1/(2y(5-3y)) = (A(5-3y)+2By)/(2y(5-3y))#

#y(2B-3A) +5A=1#

#{(2B-3A=0),(5A=1):}#

#{(A=1/5),(B=3/10):}#

#int dy/(2y(5-3y)) = 1/10int dy/y+3/10int dy/(5-3y)#

#int dy/(2y(5-3y)) = 1/10(int dy/y-int (d(5-3y))/(5-3y))#

#int dy/(2y(5-3y)) = 1/10(lnabsy-lnabs(5-3y)) + C#

#int dy/(2y(5-3y)) =-1/10lnabs(5/y-3) + C#

Substituting in #(1)#:

#x = -1/10lnabs(5/y-3) + C#

#-10x+C =lnabs(5/y-3) #

#ce^(-10x) = 5/y-3#

#5/y =3+ce^(-10x)#

#y =5/(3+ce^(-10x))#

For #x=0# the initial condition is #y(0)=2#, so:

#2=5/(3+c)#

#c=-1/2#

Then, the necessary resolution is:

#y(x) = 10/(6-e^(-10x))#

Elijah Abram · Answer

undefined The solution to the given differential equation $ \frac{dy}{dx} = 2y(5 - 3y) $ with the initial condition $ y(0) = 2 $ is:

$$ y(x) = \frac{5e^{10x}}{3e^{10x} + 1} $$