Find out the answer to "What might be a solution to the differential..."

Helen Lewis

Answered question

2022-01-20

What might be a solution to the differential equation of the form

x y = c \frac{y}{y + d}

where

y = y (x)

and c,d are constants? I am supposed to simply state a solution to this, but I don;t think it is all that obvious.

maul124uk

Beginner2022-01-20Added 35 answers

The only thing I can think of is to try to obtain a solution as a power series. First of all, the change of variable

y = d z

changes the equation into

x z^{″} = \frac{α z}{z + 1}, α = \frac{c}{d}

.
Look for a solution of the form

z (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots

It is easy tio check that the only possible value for

a_{0}

a_{0} = 0

. Then

x z^{″} = 2 a_{2} x + 6 a_{3} x^{2} + 12 a_{4} x^{3} + \dots

\frac{α z}{z + 1} = α a_{1} x + α (a_{2} - a_{1}^{2}) x^{2} + α (a_{1}^{3} - 2 a_{1} a_{2} + a_{3}) x^{3} + \dots

Equating coefficients of equal powers, we can find an expression for

a_{n}

in terms of

a_{1}, \dots, a_{n - 1}

a_{2} = \frac{α a_{1}}{2}

a_{3} = \frac{α}{6} (a_{2} - a_{1}^{2})

a_{4} = \frac{α}{12} (a_{1}^{3} - 2 a_{1} a_{2} + a_{3})

\dots = \dots

The value of

a_{1} = z^{'} (0)

is unrestricted. Of course, to prove that z is in fact a solution, one must show that the series has a positive radius of convergence.

Thomas White

Beginner2022-01-21Added 40 answers

Try

y = d \exp (x) (d \neq 0)

. Then we have

x (d \exp (x)) = c \frac{d \exp (x)}{d \exp (x) + d}

which after simplification gives an implicit solution:

y = c \frac{\exp (x)}{x (\exp (x) + 1)}

.
Which can be further simplied as :

y = c \frac{\exp (x)}{y + x} \Rightarrow y^{2} + y x = c \exp (x)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?