Question

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

Answer 1

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

Explanation:

We have:

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

This is a First Order Separable ODE, so we can "separate the variables" to get

`int \ 1/(y(5-3y)) \ dy = int \ 2 \ dx`

The RHS is trivial, and the LHS can be integrated by decomposing the integrand into partial fractions:

`1/(y(5-3y)) -= A/y+ B/(5-3y)`
`" " = (A(5-3y)+By)/(y(5-3y))`

Leading to the identity:

`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`

So 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 integrating we get:

`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`

So 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)`

Answer 2

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

Explanation:

This is a separable differential equation:

`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`

The required solution is then:

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